The program of the elective course "Conversion of numerical and alphabetic expressions"
Explanatory note
AT last years the quality of school mathematical education is checked with the help of KIMs, the main part of the tasks of which are offered in a test form. This form of testing differs from the classical examination paper and requires specific preparation. A feature of testing in the form that has developed to date is the need to respond to a large number of questions for a limited period of time, i.e. it is required not only to answer the questions correctly, but also to do it quickly enough. Therefore, it is important for students to learn various tricks, methods that will allow them to achieve the desired result.
When solving almost any school mathematical problem, you have to do some transformations. Often, its complexity is completely determined by the degree of complexity and the amount of transformations that need to be performed. It is not uncommon for a student to be unable to solve a problem, not because he does not know how it is solved, but because he cannot, without errors, perform all the necessary transformations and calculations in the allotted time.
Examples for converting numerical expressions are not important in themselves, but as a means of developing the technique of converting. From year to year schooling the concept of a number expands from natural to real and, in high school, transformations of power, logarithmic and trigonometric expressions. This material is quite difficult to study, as it contains many formulas and conversion rules.
To simplify an expression, perform the required actions, or calculate the value of an expression, you need to know in which direction you should “move” along the path of transformations that lead to the shortest “route” to the correct answer. The choice of a rational path largely depends on the possession of the entire amount of information about the ways of transforming expressions.
In high school, there is a need to systematize and deepen knowledge and practical skills in working with numerical expressions. As statistics show, about 30% of errors made when entering universities are of a computational nature. Therefore, when considering relevant topics in the middle level and when repeated in the senior level, it is necessary to pay more attention to the development of computational skills in schoolchildren.
Therefore, to help teachers teaching in the 11th grade specialized school can be suggested elective course"Converting numeric and alphabetic expressions to school course mathematics."
Classes:== 11
Type of elective course:
systematizing, generalizing and deepening course.
Number of hours:
34 (per week - 1 hour)
Educational area:
mathematics
Goals and objectives of the course:
Systematization, generalization and expansion of students' knowledge about numbers and actions with them; - formation of interest in the computational process; - development of independence, creative thinking and cognitive interest students; - adaptation of students to the new rules for entering universities.
Organization of the course
The elective course "Conversion of Numerical and Letter Expressions" expands and deepens the basic program in mathematics in high school and is designed for 11th grade. The proposed course aims to develop computational skills and sharpness of thinking. The course is built according to the classical lesson scheme, with an emphasis on workshops. It is designed for students with a high or average level of mathematical training and is designed to help them prepare for admission to universities, to contribute to the continuation of a serious mathematical education.
Planned results:
Knowledge of the classification of numbers;
Improving the skills and abilities of quick counting;
Ability to use mathematical apparatus in solving various problems;
Development logical thinking, contributing to the continuation of a serious mathematical education.
The content of the elective subject "Conversion of numerical and alphabetic expressions"
Whole numbers (4h): Number row. Fundamental theorem of arithmetic. NOD and NOC. divisibility signs. Method of mathematical induction.
Rational numbers (2h): Definition of a rational number. Basic property of a fraction. Abbreviated multiplication formulas. Definition of a periodic fraction. The rule for converting from a decimal periodic fraction to an ordinary.
Irrational numbers. Radicals. Degrees. Logarithms (6h): Definition of an irrational number. Proof of the irrationality of a number. Getting rid of irrationality in the denominator. Real numbers. Degree properties. properties of arithmetic root nth degree. Definition of a logarithm. Properties of logarithms.
Trigonometric functions (4h): Number circle. Numerical values of trigonometric functions of basic angles. Converting an angle from degrees to radians and vice versa. Basic trigonometric formulas. Casting formulas. Reverse trigonometric functions. Trigonometric operations on arc functions. Basic relationships between arc functions.
Complex numbers (2h): The concept of a complex number. Actions with complex numbers. Trigonometric and exponential forms of a complex number.
Intermediate testing (2h)
Comparison of numerical expressions (4h): Numerical inequalities on the set real numbers. Properties numerical inequalities. Supporting inequalities. Methods for proving numerical inequalities.
Letter expressions (8h): Rules for transforming expressions with variables: polynomials; algebraic fractions; irrational expressions; trigonometric and other expressions. Proofs of identities and inequalities. Simplifying expressions.
Educational and thematic plan
The plan is for 34 hours. It is compiled taking into account the topic of the diploma, therefore two separate parts are considered: numerical and literal expressions. At the discretion of the teacher, alphabetic expressions can be considered together with numerical ones in the relevant topics.
| № | Topic of the lesson | Number of hours |
| 1.1 | Whole numbers | 2 |
| 1.2 | Method of mathematical induction | 2 |
| 2.1 | Rational numbers | 1 |
| 2.2 | Decimal Periodic Fractions | 1 |
| 3.1 | Irrational numbers | 2 |
| 3.2 | Roots and degrees | 2 |
| 3.3 | Logarithms | 2 |
| 4.1 | Trigonometric functions | 2 |
| 4.2 | Inverse trigonometric functions | 2 |
| 5 | Complex numbers | 2 |
| Test on the topic "Numeric expressions" | 2 | |
| 6 | Comparing Numeric Expressions | 4 |
| 7.1 | Converting expressions with radicals | 2 |
| 7.2 | Converting power and logarithmic expressions | 2 |
| 7.3 | Converting trigonometric expressions | 2 |
| Final test | 2 | |
| Total | 34 |
Expressions, expression conversion
Power expressions (expressions with powers) and their transformation
In this article, we will talk about transforming expressions with powers. First, we will focus on the transformations that are performed with expressions of any kind, including power expressions, such as opening brackets, reducing similar terms. And then we will analyze the transformations inherent specifically in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.
Page navigation.
What are Power Expressions?
The term "power expressions" is practically not found in school textbooks of mathematics, but it often appears in collections of problems, especially designed to prepare for the Unified State Examination and the OGE, for example,. After analyzing tasks in which it is required to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing degrees in their entries. Therefore, for yourself, you can take the following definition:
Definition.
Power expressions are expressions containing powers.
Let's bring examples of power expressions. Moreover, we will present them according to how the development of views on the degree of natural indicator up to the real exponent.
As you know, first there is an acquaintance with the degree of a number with a natural exponent, at this stage the first simplest power expressions of the type 3 2 , 7 5 +1 , (2+1) 5 , (−0,1) 4 , 3 a 2 −a+a 2 , x 3−1 , (a 2) 3 etc.
A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with negative integer powers, like the following: 3 −2, 
, a −2 +2 b −3 + c 2 .
In the senior classes, they return to the degrees again. There, a degree with a rational exponent is introduced, which leads to the appearance of the corresponding power expressions: 
, , 
etc. Finally, degrees with irrational exponents and expressions containing them are considered: , .
The matter is not limited to the listed power expressions: further the variable penetrates into the exponent, and there are, for example, such expressions 2 x 2 +1 or 
. And after getting acquainted with, expressions with powers and logarithms begin to appear, for example, x 2 lgx −5 x lgx.
So, we figured out the question of what are power expressions. Next, we will learn how to transform them.
The main types of transformations of power expressions
With power expressions, you can perform any of the basic identity transformations of expressions. For example, you can expand brackets, replace numeric expressions with their values, add like terms, and so on. Naturally, in this case it is necessary to follow the accepted procedure for performing actions. Let's give examples.
Example.
Calculate the value of the power expression 2 3 ·(4 2 −12) .
Decision.
According to the order of the actions, we first perform the actions in brackets. There, firstly, we replace the power of 4 2 with its value 16 (see if necessary), and secondly, we calculate the difference 16−12=4 . We have 2 3 (4 2 −12)=2 3 (16−12)=2 3 4.
In the resulting expression, we replace the power of 2 3 with its value 8 , after which we calculate the product 8·4=32 . This is the desired value.
So, 2 3 (4 2 −12)=2 3 (16−12)=2 3 4=8 4=32.
Answer:
2 3 (4 2 −12)=32 .
Example.
Simplify Power Expressions 3 a 4 b −7 −1+2 a 4 b −7.
Decision.
Obviously, this expression contains similar terms 3 · a 4 · b − 7 and 2 · a 4 · b − 7 , and we can reduce them: .
Answer:
3 a 4 b −7 −1+2 a 4 b −7 =5 a 4 b −7 −1.
Example.
Express an expression with powers as a product.
Decision.
To cope with the task allows the representation of the number 9 as a power of 3 2 and the subsequent use of the abbreviated multiplication formula, the difference of squares: 
Answer:
There are also a number of identical transformations inherent in power expressions. Next, we will analyze them.
Working with base and exponent
There are degrees, in the basis and / or indicator of which are not just numbers or variables, but some expressions. As an example, let's write (2+0.3 7) 5−3.7 and (a (a+1)−a 2) 2 (x+1) .
When working with such expressions, it is possible to replace both the expression in the base of the degree and the expression in the indicator with an identically equal expression on the DPV of its variables. In other words, according to the rules known to us, we can separately convert the base of the degree, and separately - the indicator. It is clear that as a result of this transformation, an expression is obtained that is identically equal to the original one.
Such transformations allow us to simplify expressions with powers or achieve other goals we need. For example, in the power expression (2+0.3 7) 5−3.7 mentioned above, you can perform operations with numbers in the base and exponent, which will allow you to go to the power of 4.1 1.3. And after opening the brackets and bringing like terms in the base of the degree (a·(a+1)−a 2) 2·(x+1) we get a power expression of a simpler form a 2·(x+1) .
Using Power Properties
One of the main tools for transforming expressions with powers is equalities that reflect . Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following power properties hold:
- a r a s =a r+s ;
- a r:a s =a r−s ;
- (a b) r = a r b r ;
- (a:b) r =a r:b r ;
- (a r) s =a r s .
Note that for natural, integer, and positive exponents, restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n, the equality a m a n =a m+n is true not only for positive a , but also for negative ones, and for a=0 .
At school, the main attention in the transformation of power expressions is focused precisely on the ability to choose the appropriate property and apply it correctly. In this case, the bases of the degrees are usually positive, which allows you to use the properties of the degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of acceptable values of variables is usually such that the bases take only positive values on it, which allows you to freely use the properties of degrees. In general, you need to constantly ask yourself whether it is possible to apply any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the ODZ and other troubles. These points are discussed in detail and with examples in the article transformation of expressions using the properties of degrees. Here we confine ourselves to a few simple examples.
Example.
Express the expression a 2.5 ·(a 2) −3:a −5.5 as a power with base a .
Decision.
First, we transform the second factor (a 2) −3 by the property of raising a power to a power: (a 2) −3 =a 2 (−3) =a −6. In this case, the initial power expression will take the form a 2.5 ·a −6:a −5.5 . Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 a -6:a -5.5 =
a 2.5−6:a−5.5 =a−3.5:a−5.5 =
a −3.5−(−5.5) =a 2 .
Answer:
a 2.5 (a 2) -3:a -5.5 \u003d a 2.
Power properties are used when transforming power expressions both from left to right and from right to left.
Example.
Find the value of the power expression.
Decision.
Equality (a·b) r =a r ·b r , applied from right to left, allows you to go from the original expression to the product of the form and further. And when multiplying powers with the same grounds indicators add up: 
.
It was possible to perform the transformation of the original expression in another way: 
Answer:
.
Example.
Given a power expression a 1.5 −a 0.5 −6 , enter a new variable t=a 0.5 .
Decision.
The degree a 1.5 can be represented as a 0.5 3 and further on the basis of the property of the degree in the degree (a r) s =a r s applied from right to left, convert it to the form (a 0.5) 3 . Thus, a 1.5 -a 0.5 -6=(a 0.5) 3 -a 0.5 -6. Now it is easy to introduce a new variable t=a 0.5 , we get t 3 −t−6 .
Answer:
t 3 −t−6 .
Converting fractions containing powers
Power expressions can contain fractions with powers or represent such fractions. Any of the basic fraction transformations that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain degrees can be reduced, reduced to a new denominator, work separately with their numerator and separately with the denominator, etc. To illustrate the above words, consider the solutions of several examples.
Example.
Simplify Power Expression 
.
Decision.
This power expression is a fraction. Let's work with its numerator and denominator. In the numerator, we open the brackets and simplify the expression obtained after that using the properties of powers, and in the denominator we present similar terms: 
And we also change the sign of the denominator by placing a minus in front of the fraction: 
.
Answer:
.
Reducing fractions containing powers to a new denominator is carried out similarly to reducing rational fractions to a new denominator. At the same time, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that reduction to a new denominator can lead to a narrowing of the DPV. To prevent this from happening, it is necessary that the additional factor does not vanish for any values of the variables from the ODZ variables for the original expression.
Example.
Bring the fractions to a new denominator: a) to the denominator a, b) 
to the denominator.
Decision.
a) In this case, it is quite easy to figure out what additional factor helps to achieve the desired result. This is a multiplier a 0.3, since a 0.7 a 0.3 = a 0.7+0.3 = a . Note that in the range of acceptable values of the variable a (this is the set of all positive real numbers), the degree a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of the given fraction by this additional factor: 
b) Looking more closely at the denominator, we find that 
and multiplying this expression by will give the sum of cubes and , that is, . And this is the new denominator to which we need to bring the original fraction.
So we found an additional factor . The expression does not vanish on the range of acceptable values of the variables x and y, therefore, we can multiply the numerator and denominator of the fraction by it: 
Answer:
a) 
, b) 
.
There is also nothing new in the reduction of fractions containing degrees: the numerator and denominator are represented as a certain number of factors, and the same factors of the numerator and denominator are reduced.
Example.
Reduce the fraction: a) 
, b).
Decision.
a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which equals 15. Also, obviously, you can reduce by x 0.5 +1 and by 
. Here's what we have: 
b) In this case, the same factors in the numerator and denominator are not immediately visible. To get them, you have to perform preliminary transformations. In this case, they consist in decomposing the denominator into factors according to the difference of squares formula: 
Answer:
a) 
b) 
.
Reducing fractions to a new denominator and reducing fractions is mainly used to perform operations on fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are reduced to a common denominator, after which the numerators are added (subtracted), and the denominator remains the same. The result is a fraction whose numerator is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by its reciprocal.
Example.
Follow the steps 
.
Decision.
First, we subtract the fractions in brackets. To do this, we bring them to a common denominator, which is 
, then subtract the numerators: 
Now we multiply fractions: 
Obviously, a reduction by the power x 1/2 is possible, after which we have 
.
You can also simplify the power expression in the denominator by using the difference of squares formula: 
.
Answer:
Example.
Simplify Power Expression 
.
Decision.
Obviously, this fraction can be reduced by (x 2.7 +1) 2, this gives the fraction 
. It is clear that something else needs to be done with the powers of x. To do this, we convert the resulting fraction into a product. This gives us the opportunity to use the property of dividing powers with the same bases: 
. And at the end of the process, we pass from the last product to the fraction.
Answer:
.
And we add that it is possible and in many cases desirable to transfer factors with negative exponents from the numerator to the denominator or from the denominator to the numerator by changing the sign of the exponent. Such transformations often simplify further actions. For example, a power expression can be replaced by .
Converting expressions with roots and powers
Often in expressions in which some transformations are required, along with degrees with fractional exponents, there are also roots. To convert such an expression to the desired form, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with degrees, they usually move from roots to degrees. However, it is advisable to carry out such a transition when the ODZ of variables for the original expression allows you to replace the roots with degrees without the need to access the module or split the ODZ into several intervals (we discussed this in detail in the article, the transition from roots to powers and vice versa After getting acquainted with the degree with a rational exponent a degree with an irrational indicator is introduced, which makes it possible to speak of a degree with an arbitrary real indicator.At this stage, the school begins to study exponential function, which is analytically given by the degree, in the basis of which there is a number, and in the indicator - a variable. So we are faced with exponential expressions containing numbers in the base of the degree, and in the exponent - expressions with variables, and naturally the need arises to perform transformations of such expressions.
It should be said that the transformation of expressions of the indicated type usually has to be performed when solving exponential equations and exponential inequalities , and these transformations are quite simple. In the vast majority of cases, they are based on the properties of the degree and are aimed mostly at introducing a new variable in the future. The equation will allow us to demonstrate them 5 2 x+1 −3 5 x 7 x −14 7 2 x−1 =0.
First, the exponents, in whose exponents the sum of some variable (or expression with variables) and a number, is found, are replaced by products. This applies to the first and last terms of the expression on the left side:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 =0,
5 5 2 x −3 5 x 7 x −2 7 2 x =0.
Next, both parts of the equality are divided by the expression 7 2 x , which takes only positive values on the ODV of the x variable for the original equation (this is a standard technique for solving equations of this kind, we are not talking about it now, so focus on subsequent transformations of expressions with powers ): 
Now fractions with powers are cancelled, which gives 
.
Finally, the ratio of powers with the same exponents is replaced by powers of ratios, which leads to the equation 
, which is equivalent to 
. The transformations made allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of the quadratic equation
TOPIC OF ELECTIVE SUBJECT
CONVERSION OF NUMERIC AND LETTER EXPRESSIONS
Quantity 34 hours
higher mathematics teacher
MOU "Secondary School No. 51"
Saratov, 2008
ELECTIVE SUBJECT PROGRAM
"CONVERSION OF NUMERICAL AND LETTER EXPRESSIONS"
Explanatory note
In recent years, final exams in schools, as well as entrance exams in universities, are conducted with the help of tests. This form of testing is different from the classic exam and requires specific preparation. A feature of testing in the form that has developed to date is the need to answer a large number of questions in a limited period of time, that is, it is required not only to answer the questions posed, but also to do it quickly. Therefore, it is important to master various techniques, methods that allow you to achieve the desired result.
When solving almost any school problem, you have to make some transformations. Often, its complexity is completely determined by the degree of complexity and the amount of transformations that need to be performed. It is not uncommon for a student to be unable to solve a problem, not because he does not know how it is solved, but because he cannot make all the necessary transformations and calculations in a reasonable time without errors.
The elective course "Transformation of Numerical and Letter Expressions" expands and deepens the basic program in mathematics in high school and is designed for study in grade 11. The proposed course aims to develop computational skills and sharpness of thinking. The course is designed for students with a high or average level of mathematical training and is designed to help them prepare for admission to universities, to contribute to the continuation of a serious mathematical education.
Goals and objectives:
Systematization, generalization and expansion of students' knowledge about numbers and actions with them;
Development of independence, creative thinking and cognitive interest of students;
Formation of interest in the computing process;
Adaptation of students to the new rules for entering universities.
Expected results:
Knowledge of the classification of numbers;
Improving the skills and abilities of quick counting;
Ability to use mathematical apparatus in solving various problems;
Educational and thematic plan
The plan is for 34 hours. It is compiled taking into account the topic of the diploma, so two separate parts are considered: numerical and alphabetic expressions. At the discretion of the teacher, alphabetic expressions can be considered together with numerical ones in the relevant topics.
Number of hours |
||
Numeric expressions Whole numbers Method of mathematical induction Rational numbers Decimal Periodic Fractions Irrational numbers Roots and degrees Logarithms Trigonometric functions Inverse trigonometric functions Complex numbers Test on the topic "Numeric expressions" Comparing Numeric Expressions Literal expressions Converting expressions with radicals Power Expression Transformation Converting Logarithmic Expressions Converting trigonometric expressions Final test |
Whole numbers (4h)
Number row. Fundamental theorem of arithmetic. NOD and NOC. divisibility signs. Method of mathematical induction.
Rational numbers (2h)
Definition of a rational number. Basic property of a fraction. Abbreviated multiplication formulas. Definition of a periodic fraction. The rule for converting from a decimal periodic fraction to an ordinary.
Irrational numbers. Radicals. Degrees. Logarithms (6h)
Definition of an irrational number. Proof of the irrationality of a number. Getting rid of irrationality in the denominator. Real numbers. Degree properties. Properties arithmetic root nth degree. Definition of a logarithm. Properties of logarithms.
Trigonometric functions (4h)
Number circle. Numerical values of trigonometric functions of basic angles. Converting an angle from degrees to radians and vice versa. Basic trigonometric formulas. Casting formulas. Inverse trigonometric functions. Trigonometric operations on arc functions. Basic relationships between arc functions.
Complex numbers (2h)
The concept of a complex number. Operations with complex numbers. Trigonometric and exponential forms of a complex number.
Intermediate testing (2h)
Comparison of numerical expressions (4 hours)
Numerical inequalities on the set of real numbers. Properties of numerical inequalities. Supporting inequalities. Methods for proving numerical inequalities.
Letter expressions (8h)
Rules for transforming expressions with variables: polynomials; algebraic fractions; irrational expressions; trigonometric and other expressions. Proofs of identities and inequalities. Simplifying expressions.
1 part of the elective subject: "Numeric expressions"
ACTIVITY 1(2 hours)
Lesson topic: Whole numbers
Lesson Objectives: Generalize and systematize students' knowledge about numbers; recall the concepts of GCD and NOC; expand knowledge about the signs of divisibility; consider problems solved in integers.
During the classes
I. Introductory lecture.
Number classification:
Integers;
Whole numbers;
Rational numbers;
Real numbers;
Complex numbers.
Acquaintance with the number series at school begins with the concept of a natural number. The numbers used in counting objects are called natural. The set of natural numbers is denoted by N. Natural numbers are divided into prime and composite. Prime numbers have only two divisors one and the number itself, while composite numbers have more than two divisors. Fundamental theorem of arithmetic states: "Any natural number greater than 1 can be represented as a product of prime numbers (not necessarily different ones), and, moreover, in a unique way (up to the order of factors)."
Two more important arithmetic concepts are associated with natural numbers: the greatest common divisor (GCD) and the least common multiple (LCM). Each of these concepts actually defines itself. The solution of many problems is facilitated by the signs of divisibility, which must be remembered.
Sign of divisibility by 2 . A number is divisible by 2 if its last digit is even or o.
Divisibility by 4 sign . A number is divisible by 4 if the last two digits are zeros or form a number divisible by 4.
Sign of divisibility by 8. A number is divisible by 8 if its last three digits are zeros or form a number divisible by 8.
Divisibility criteria for 3 and 9. Only those numbers are divisible by 3 for which the sum of the digits is divisible by 3; by 9 - only those in which the sum of the digits is divisible by 9.
Sign of divisibility by 6. A number is divisible by 6 if it is divisible by both 2 and 3.
Sign of divisibility by 5 . Divisible by 5 are numbers whose last digit is 0 or 5.
Sign of divisibility by 25. Divisible by 25 are numbers whose last two digits are zeros or form a number that is divisible by 25.
Signs of divisibility by 10,100,1000. Only those numbers whose last digit is 0 are divisible by 10, only those numbers whose last two digits are 0 are divisible by 100, only those numbers whose last three digits are 0 are divisible by 1000.
Sign of divisibility by 11 . Only those numbers are divisible by 11 in which the sum of the digits occupying odd places is either equal to the sum of the digits occupying even places, or differs from it by a number divisible by 11.
In the first lesson, we will look at natural and integer numbers. whole numbers are natural numbers, their opposite numbers and zero. The set of integers is denoted by Z.
II. Problem solving.
EXAMPLE 1. Factorize: a) 899; b) 1000027.
Solution: a) ;
b) EXAMPLE 2. Find the GCD of the numbers 2585 and 7975.
Solution: Let's use the Euclid algorithm:
If https://pandia.ru/text/78/342/images/image004_155.gif" width="167" height="29 src=">;
https://pandia.ru/text/78/342/images/image006_127.gif" width="88" height="29 src=">.gif" width="16" height="29">
220 |165 -
165|55 -
Answer: gcd(2585,7975) = 55.
EXAMPLE 3 Calculate:
Solution: = 1987100011989. The second product is equal to the same value. Therefore, the difference is 0.
EXAMPLE 4. Find GCD and LCM numbers a) 5544 and 1404; b) 198, 504 and 780.
Answers: a) 36; 49896; b) 6; 360360.
EXAMPLE 5. Find the quotient and remainder when dividing
a) 5 to 7; https://pandia.ru/text/78/342/images/image013_75.gif" width="109" height="20 src=">;
c) -529 to (-23); https://pandia.ru/text/78/342/images/image015_72.gif" width="157" height="28 src=">;
e) 256 to (-15); https://pandia.ru/text/78/342/images/image017_68.gif" width="101" height="23">
Solution: https://pandia.ru/text/78/342/images/image020_64.gif" width="123 height=28" height="28">.
b) 
Solution: https://pandia.ru/text/78/342/images/image024_52.gif" width="95" height="23">.
EXAMPLE 7..gif" width="67" height="27 src="> by 17.
Solution: Let's enter a record 
, which means that when divided by m, the numbers a, b, c, ... d give the same remainder.
Therefore, for any natural k, there will be
But 1989=16124+5. Means,
Answer: The remainder is 12.
EXAMPLE 8. Find the smallest natural number greater than 10, which, when divided by 24, 45, and 56, would give a remainder of 1.
Answer: LCM(24;45;56)+1=2521.
EXAMPLE 9. Find the smallest natural number that is divisible by 7, and when divided by 3, 4 and 5 gives a remainder of 1.
Answer: 301. Instruction. Among the numbers of the form 60k + 1, you need to find the smallest divisible by 7; k = 5.
EXAMPLE 10. Assign to 23 one digit on the right and on the left so that the resulting four-digit number is divisible by 9 and 11.
Answer: 6237.
EXAMPLE 11. Assign three digits to the back of the number so that the resulting number is divisible by 7, 8 and 9.
Answer: 304 or 808. Indication. The number when divided by = 789) gives a remainder of 200. Therefore, if you add 304 or 808 to it, it will be divided by 504.
EXAMPLE 12. Is it possible to rearrange the digits in a three-digit number divisible by 37 so that the resulting number is also divisible by 37?
Answer: You can. Note..gif" width="61" height="24"> is also divisible by 37. We have A = 100a + 10b + c = 37k, whence c = 37k -100a - 10b. Then B = 100b + 10c + a = 100b + k - 100a - 10b) + a \u003d 370k - 999a, that is, B is divisible by 37.
EXAMPLE 13. Find the number, when divided by which the numbers 1108, 1453, 1844 and 2281 give the same remainder.
Answer: 23. Indication. The difference of any two given numbers is divisible by the required one. This means that any common divisor of all possible data differences, other than 1, is suitable for us
EXAMPLE 14. Represent 19 as the difference of cubes of natural numbers.
EXAMPLE 15. The square of a natural number is equal to the product of four consecutive odd numbers. Find this number.
Answer: 
.
EXAMPLE 16..gif" width="115" height="27"> is not divisible by 10.
Answer: a) Direction. Having grouped the first and last terms, the second and penultimate, etc., use the formula for the sum of cubes.
b) Indication..gif" width="120" height="20">.
4) Find all pairs of natural numbers whose GCD is 5 and LCM is 105.
Answer: 5, 105 or 15, 35.
ACTIVITY 2(2 hours)
Lesson topic: Method of mathematical induction.
The purpose of the lesson: Consider mathematical statements requiring proof; introduce students to the method of mathematical induction; develop logical thinking.
During the classes
I. Checking homework.
II. Explanation of new material.
In the school mathematics course, along with the tasks “Find the value of the expression”, there are tasks of the form: “Prove equality”. One of the most universal methods for proving mathematical statements in which the words “for an arbitrary natural n” appear is the method of complete mathematical induction.
A proof using this method always consists of three steps:
1) Basis of induction. The validity of the statement for n = 1 is checked.
In some cases, to start the induction, you have to check several
initial values.
2) Assumption of induction. The statement is assumed to be true for any
3) Inductive step. We prove the validity of the assertion for
Thus, starting from n = 1, on the basis of the proven inductive step, we obtain the validity of the assertion being proved for
n =2, 3,…t. e. for any n.
Let's look at a few examples.
EXAMPLE 1: Prove that for any natural n the number 
is divisible by 7.
Proof: Denote 
.
Step 1..gif" width="143" height="37 src="> is divisible by 7.
Step 3..gif" width="600" height="88">
The last number is divisible by 7 because it is the difference between two integers divisible by 7.
EXAMPLE 2: Prove equality https://pandia.ru/text/78/342/images/image047_31.gif" width="240" height="36 src=">
https://pandia.ru/text/78/342/images/image049_34.gif" width="157" height="47"> is obtained from 
replacing n with k = 1.
III. Problem solving
In the first lesson, from the tasks below (No. 1-3), several are selected for solution at the discretion of the teacher for analysis on the board. The second lesson deals with № 4.5; held independent work from #1-3; No. 6 is offered as an additional one, with a mandatory decision on the board.
1) Prove that a) is divisible by 83;
b) is divisible by 13;
c) is divisible by 20801.
2) Prove that for any natural n:
a) 
is divisible by 120;
b) 
is divisible by 27;
in) 
divisible by 84;
G) 
is divisible by 169;
e) 
is divisible by 8;
f) is divisible by 8;
g) is divisible by 16;
h) 
divisible by 49;
and) 
is divisible by 41;
to) 
is divisible by 23;
l) 
is divisible by 13;
m) 
divided by .
3) Prove that:
G) 
;
4) Output the sum formula https://pandia.ru/text/78/342/images/image073_23.gif" width="187" height="20">.
6) Prove that the sum of the members of each row of the table
…………….
is equal to the square of an odd number whose number in a row is equal to the row number from the beginning of the table.
Answers and instructions.
1) Let's use the entry introduced in example 4 of the previous lesson.
a) . Hence divisible by 83 
.
b) Because 
, then ;
. Hence, 
.
c) Since , it is necessary to prove that the given number is divisible by 11, 31 and 61..gif" width="120" height="32 src=">. Divisibility by 11 and 31 is proved similarly.
2) a) Let us prove that this expression is divisible by 3, 8, 5. Divisibility by 3 follows from the fact that 
, and out of three consecutive natural numbers, one is divisible by 3..gif" width="72" height="20 src=">.gif" width="75" height="20 src=">. To check divisibility by 5, it is enough to consider the values n=0,1,2,3,4.
A literal expression (or expression with variables) is a mathematical expression that consists of numbers, letters, and signs of mathematical operations. For example, the following expression is literal:
a+b+4
Using literal expressions, you can write down laws, formulas, equations, and functions. The ability to manipulate literal expressions is the key to a good knowledge of algebra and higher mathematics.
Any serious problem in mathematics comes down to solving equations. And to be able to solve equations, you need to be able to work with literal expressions.
To work with literal expressions, you need to study basic arithmetic well: addition, subtraction, multiplication, division, basic laws of mathematics, fractions, operations with fractions, proportions. And not just to study, but to understand thoroughly.
Lesson contentVariables
Letters that are contained in literal expressions are called variables. For example, in the expression a+b+ 4 variables are letters a and b. If instead of these variables we substitute any numbers, then the literal expression a+b+ 4 will turn to numeric expression, whose value can be found.
Numbers that are substituted for variables are called variable values. For example, let's change the values of the variables a and b. Use the equals sign to change values
a = 2, b = 3
We have changed the values of the variables a and b. variable a assigned a value 2 , variable b assigned a value 3 . As a result, the literal expression a+b+4 converts to a normal numeric expression 2+3+4 whose value can be found:
When variables are multiplied, they are written together. For example, the entry ab means the same as the entry a x b. If we substitute instead of variables a and b numbers 2 and 3 , then we get 6
Together, you can also write the multiplication of a number by an expression in brackets. For example, instead of a×(b + c) can be written a(b + c). Applying the distributive law of multiplication, we obtain a(b + c)=ab+ac.
Odds
In literal expressions, you can often find a notation in which a number and a variable are written together, for example 3a. In fact, this is a shorthand for multiplying the number 3 by a variable. a and this entry looks like 3×a .
In other words, the expression 3a is the product of the number 3 and the variable a. Number 3 in this work is called coefficient. This coefficient shows how many times the variable will be increased a. This expression can be read as " a three times or three times a", or "increment the value of the variable a three times", but most often read as "three a«
For example, if the variable a is equal to 5 , then the value of the expression 3a will be equal to 15.
3 x 5 = 15
talking plain language, the coefficient is the number that comes before the letter (before the variable).
There can be several letters, for example 5abc. Here the coefficient is the number 5 . This coefficient shows that the product of variables abc increases five times. This expression can be read as " abc five times" or "increase the value of the expression abc five times" or "five abc «.
If instead of variables abc substitute the numbers 2, 3 and 4, then the value of the expression 5abc will be equal to 120
5 x 2 x 3 x 4 = 120
You can mentally imagine how the numbers 2, 3 and 4 were first multiplied, and the resulting value increased five times:
The sign of the coefficient refers only to the coefficient, and does not apply to variables.
Consider the expression −6b. Minus in front of the coefficient 6 , applies only to the coefficient 6 , and does not apply to the variable b. Understanding this fact will allow you not to make mistakes in the future with signs.
Find the value of the expression −6b at b = 3.
−6b −6×b. For clarity, we write the expression −6b in expanded form and substitute the value of the variable b
−6b = −6 × b = −6 × 3 = −18
Example 2 Find the value of an expression −6b at b = −5
Let's write the expression −6b in expanded form
−6b = −6 × b = −6 × (−5) = 30
Example 3 Find the value of an expression −5a+b at a = 3 and b = 2
−5a+b is the short form for −5 × a + b, therefore, for clarity, we write the expression −5×a+b in expanded form and substitute the values of the variables a and b
−5a + b = −5 × a + b = −5 × 3 + 2 = −15 + 2 = −13
Sometimes letters are written without a coefficient, for example a or ab. In this case, the coefficient is one:
but the unit is traditionally not written down, so they just write a or ab
If there is a minus before the letter, then the coefficient is a number −1 . For example, the expression -a actually looks like −1a. This is the product of minus one and the variable a. It came out like this:
−1 × a = −1a
Here lies a small trick. In the expression -a minus before variable a actually refers to the "invisible unit" and not the variable a. Therefore, when solving problems, you should be careful.
For example, given the expression -a and we are asked to find its value at a = 2, then at school we substituted a deuce instead of a variable a and get an answer −2 , not really focusing on how it turned out. In fact, there was a multiplication of minus one by a positive number 2
-a = -1 × a
−1 × a = −1 × 2 = −2
If an expression is given -a and it is required to find its value at a = −2, then we substitute −2 instead of a variable a
-a = -1 × a
−1 × a = −1 × (−2) = 2
In order to avoid mistakes, at first invisible units can be written explicitly.
Example 4 Find the value of an expression abc at a=2 , b=3 and c=4
Expression abc 1×a×b×c. For clarity, we write the expression abc a , b and c
1 x a x b x c = 1 x 2 x 3 x 4 = 24
Example 5 Find the value of an expression abc at a=−2 , b=−3 and c=−4
Let's write the expression abc in expanded form and substitute the values of the variables a , b and c
1 × a × b × c = 1 × (−2) × (−3) × (−4) = −24
Example 6 Find the value of an expression − abc at a=3 , b=5 and c=7
Expression − abc is the short form for −1×a×b×c. For clarity, we write the expression − abc in expanded form and substitute the values of the variables a , b and c
−abc = −1 × a × b × c = −1 × 3 × 5 × 7 = −105
Example 7 Find the value of an expression − abc at a=−2 , b=−4 and c=−3
Let's write the expression − abc expanded:
−abc = −1 × a × b × c
Substitute the value of the variables a , b and c
−abc = −1 × a × b × c = −1 × (−2) × (−4) × (−3) = 24
How to determine the coefficient
Sometimes it is required to solve a problem in which it is required to determine the coefficient of an expression. In principle, this task is very simple. It is enough to be able to correctly multiply numbers.
To determine the coefficient in an expression, you need to separately multiply the numbers included in this expression, and separately multiply the letters. The resulting numerical factor will be the coefficient.
Example 1 7m×5a×(−3)×n
The expression consists of several factors. This can be clearly seen if the expression is written in expanded form. That is, works 7m and 5a write in the form 7×m and 5×a
7 × m × 5 × a × (−3) × n
We apply the associative law of multiplication, which allows us to multiply factors in any order. Namely, separately multiply the numbers and separately multiply the letters (variables):
−3 × 7 × 5 × m × a × n = −105man
The coefficient is −105 . After completion, the letter part is preferably arranged in alphabetical order:
−105 am
Example 2 Determine the coefficient in the expression: −a×(−3)×2
−a × (−3) × 2 = −3 × 2 × (−a) = −6 × (−a) = 6a
The coefficient is 6.
Example 3 Determine the coefficient in the expression: 
Let's multiply numbers and letters separately:
The coefficient is −1. Please note that the unit is not recorded, since the coefficient 1 is usually not recorded.
These seemingly simple tasks can play a very cruel joke with us. It often turns out that the sign of the coefficient is set incorrectly: either a minus is omitted or, on the contrary, it is set in vain. To avoid these annoying mistakes, it must be studied at a good level.
Terms in literal expressions
When you add several numbers, you get the sum of those numbers. Numbers that add up are called terms. There can be several terms, for example:
1 + 2 + 3 + 4 + 5
When an expression consists of terms, it is much easier to calculate it, since it is easier to add than to subtract. But the expression can contain not only addition, but also subtraction, for example:
1 + 2 − 3 + 4 − 5
In this expression, the numbers 3 and 5 are subtracted, not added. But nothing prevents us from replacing subtraction with addition. Then we again get an expression consisting of terms:
1 + 2 + (−3) + 4 + (−5)
It doesn't matter that the numbers -3 and -5 are now with a minus sign. The main thing is that all the numbers in this expression are connected by the addition sign, that is, the expression is a sum.
Both expressions 1 + 2 − 3 + 4 − 5 and 1 + 2 + (−3) + 4 + (−5) are equal to the same value - minus one
1 + 2 − 3 + 4 − 5 = −1
1 + 2 + (−3) + 4 + (−5) = −1
Thus, the value of the expression will not suffer from the fact that we replace subtraction with addition somewhere.
You can also replace subtraction with addition in literal expressions. For example, consider the following expression:
7a + 6b - 3c + 2d - 4s
7a + 6b + (−3c) + 2d + (−4s)
For any values of variables a, b, c, d and s expressions 7a + 6b - 3c + 2d - 4s and 7a + 6b + (−3c) + 2d + (−4s) will be equal to the same value.
You must be prepared for the fact that a teacher at school or a teacher at an institute can call terms even those numbers (or variables) that are not them.
For example, if the difference is written on the board a-b, then the teacher will not say that a is the minuend, and b- deductible. He will call both variables one common word — terms. And all because the expression of the form a-b mathematician sees how the sum a + (−b). In this case, the expression becomes a sum, and the variables a and (−b) become components.
Similar terms
Similar terms are terms that have the same letter part. For example, consider the expression 7a + 6b + 2a. Terms 7a and 2a have the same letter part - variable a. So the terms 7a and 2a are similar.
Usually, like terms are added to simplify an expression or solve an equation. This operation is called reduction of like terms.
To bring like terms, you need to add the coefficients of these terms, and multiply the result by the common letter part.
For example, we give similar terms in the expression 3a + 4a + 5a. In this case, all terms are similar. We add their coefficients and multiply the result by the common letter part - by the variable a
3a + 4a + 5a = (3 + 4 + 5)×a = 12a
Such terms are usually given in the mind and the result is recorded immediately:
3a + 4a + 5a = 12a
Also, you can argue like this:
There were 3 variables a , 4 more variables a and 5 more variables a were added to them. As a result, we got 12 variables a
Let's consider several examples of reducing similar terms. Considering that this topic is very important, at first we will write down every detail in detail. Despite the fact that everything is very simple here, most people make a lot of mistakes. Mostly due to inattention, not ignorance.
Example 1 3a + 2a + 6a + 8a
We add the coefficients in this expression and multiply the result by the common letter part:
3a + 2a + 6a + 8a=(3 + 2 + 6 + 8)× a = 19a
Construction (3 + 2 + 6 + 8) × a you can not write down, so we will immediately write down the answer
3 a + 2 a + 6 a + 8 a = 19 a
Example 2 Bring like terms in the expression 2a+a
Second term a written without a coefficient, but in fact it is preceded by a coefficient 1 , which we do not see due to the fact that it is not recorded. So the expression looks like this:
2a + 1a
Now we present similar terms. That is, we add the coefficients and multiply the result by the common letter part:
2a + 1a = (2 + 1) × a = 3a
Let's write the solution in short:
2a + a = 3a
2a+a, you can argue in another way:
Example 3 Bring like terms in the expression 2a - a
Let's replace subtraction with addition:
2a + (−a)
Second term (−a) written without a coefficient, but in fact it looks like (−1a). Coefficient −1 again invisible due to the fact that it is not recorded. So the expression looks like this:
2a + (−1a)
Now we present similar terms. We add the coefficients and multiply the result by the common letter part:
2a + (−1a) = (2 + (−1)) × a = 1a = a
Usually written shorter:
2a − a = a
Bringing like terms in the expression 2a−a You can also argue in another way:
There were 2 variables a , subtracted one variable a , as a result there was only one variable a
Example 4 Bring like terms in the expression 6a - 3a + 4a - 8a
6a − 3a + 4a − 8a = 6a + (−3a) + 4a + (−8a)
Now we present similar terms. We add the coefficients and multiply the result by the common letter part
(6 + (−3) + 4 + (−8)) × a = −1a = −a
Let's write the solution in short:
6a - 3a + 4a - 8a = -a
There are expressions that contain several different groups of similar terms. For example, 3a + 3b + 7a + 2b. For such expressions, the same rules apply as for the rest, namely, adding the coefficients and multiplying the result by the common letter part. But in order to avoid mistakes, it is convenient to underline different groups of terms with different lines.
For example, in the expression 3a + 3b + 7a + 2b those terms that contain a variable a, can be underlined with one line, and those terms that contain a variable b, can be underlined with two lines:
Now we can bring like terms. That is, add the coefficients and multiply the result by the common letter part. This must be done for both groups of terms: for terms containing a variable a and for terms containing the variable b.
3a + 3b + 7a + 2b = (3+7)×a + (3 + 2)×b = 10a + 5b
Again, we repeat, the expression is simple, and similar terms can be given in the mind:
3a + 3b + 7a + 2b = 10a + 5b
Example 5 Bring like terms in the expression 5a - 6a - 7b + b
We replace subtraction with addition where possible:
5a − 6a −7b + b = 5a + (−6a) + (−7b) + b
Underline like terms with different lines. Terms containing variables a underline with one line, and the terms containing the variables b, underlined with two lines:
Now we can bring like terms. That is, add the coefficients and multiply the result by the common letter part:
5a + (−6a) + (−7b) + b = (5 + (−6))×a + ((−7) + 1)×b = −a + (−6b)
If the expression contains ordinary numbers without alphabetic factors, then they are added separately.
Example 6 Bring like terms in the expression 4a + 3a − 5 + 2b + 7
Let's replace subtraction with addition where possible:
4a + 3a − 5 + 2b + 7 = 4a + 3a + (−5) + 2b + 7
Let us present similar terms. Numbers −5 and 7 do not have literal factors, but they are similar terms - you just need to add them up. And the term 2b will remain unchanged, since it is the only one in this expression that has a letter factor b, and there is nothing to add it with:
4a + 3a + (−5) + 2b + 7 = (4 + 3)×a + 2b + (−5) + 7 = 7a + 2b + 2
Let's write the solution in short:
4a + 3a − 5 + 2b + 7 = 7a + 2b + 2
Terms can be ordered so that those terms that have the same letter part are located in the same part of the expression.
Example 7 Bring like terms in the expression 5t+2x+3x+5t+x
Since the expression is the sum of several terms, this allows us to evaluate it in any order. Therefore, the terms containing the variable t, can be written at the beginning of the expression, and the terms containing the variable x at the end of the expression:
5t+5t+2x+3x+x
Now we can add like terms:
5t + 5t + 2x + 3x + x = (5+5)×t + (2+3+1)×x = 10t + 6x
Let's write the solution in short:
5t + 2x + 3x + 5t + x = 10t + 6x
The sum of opposite numbers is zero. This rule also works for literal expressions. If the expression contains identical terms, but with opposite signs, then you can get rid of them at the stage of reducing similar terms. In other words, just drop them from the expression because their sum is zero.
Example 8 Bring like terms in the expression 3t − 4t − 3t + 2t
Let's replace subtraction with addition where possible:
3t − 4t − 3t + 2t = 3t + (−4t) + (−3t) + 2t
Terms 3t and (−3t) are opposite. The sum of opposite terms is equal to zero. If we remove this zero from the expression, then the value of the expression will not change, so we will remove it. And we will remove it by the usual deletion of the terms 3t and (−3t)
As a result, we will have the expression (−4t) + 2t. In this expression, you can add like terms and get the final answer:
(−4t) + 2t = ((−4) + 2)×t = −2t
Let's write the solution in short:
Expression simplification
"simplify the expression" and the following is the expression to be simplified. Simplify Expression means to make it simpler and shorter.
In fact, we have already dealt with the simplification of expressions when reducing fractions. After the reduction, the fraction became shorter and easier to read.
Consider the following example. Simplify the expression.
This task can be literally understood as follows: "Do whatever you can do with this expression, but make it simpler" .
In this case, you can reduce the fraction, namely, divide the numerator and denominator of the fraction by 2:
What else can be done? You can calculate the resulting fraction. Then we get the decimal 0.5
As a result, the fraction was simplified to 0.5.
The first question to ask yourself when solving such problems should be “what can be done?” . Because there are things you can do and there are things you can't do.
Another important point to keep in mind is that the value of an expression must not change after the expression is simplified. Let's return to the expression. This expression is a division that can be performed. Having performed this division, we get the value of this expression, which is equal to 0.5
But we simplified the expression and got a new simplified expression . The value of the new simplified expression is still 0.5
But we also tried to simplify the expression by calculating it. As a result, the final answer was 0.5.
Thus, no matter how we simplify the expression, the value of the resulting expressions is still 0.5. This means that the simplification was carried out correctly at each stage. This is what we need to strive for when simplifying expressions - the meaning of the expression should not suffer from our actions.
It is often necessary to simplify literal expressions. For them, the same simplification rules apply as for numerical expressions. You can perform any valid action, as long as the value of the expression does not change.
Let's look at a few examples.
Example 1 Simplify Expression 5.21s × t × 2.5
To simplify this expression, you can multiply the numbers separately and multiply the letters separately. This task is very similar to the one we considered when we learned to determine the coefficient:
5.21s × t × 2.5 = 5.21 × 2.5 × s × t = 13.025 × st = 13.025st
So the expression 5.21s × t × 2.5 simplified to 13.025st.
Example 2 Simplify Expression −0.4×(−6.3b)×2
Second work (−6.3b) can be translated into a form understandable to us, namely, written in the form ( −6.3)×b , then separately multiply the numbers and separately multiply the letters:
− 0,4 × (−6.3b) × 2 = − 0,4 × (−6.3) × b × 2 = 5.04b
So the expression −0.4×(−6.3b)×2 simplified to 5.04b
Example 3 Simplify Expression 
Let's write this expression in more detail in order to clearly see where the numbers are and where the letters are:
Now we multiply the numbers separately and multiply the letters separately:
So the expression simplified to −abc. This solution can be written shorter:
When simplifying expressions, fractions can be reduced in the process of solving, and not at the very end, as we did with ordinary fractions. For example, if in the course of solving we come across an expression of the form , then it is not at all necessary to calculate the numerator and denominator and do something like this:
A fraction can be reduced by choosing both the factor in the numerator and the denominator and reducing these factors by their greatest common divisor. In other words, use , in which we do not describe in detail what the numerator and denominator were divided into.
For example, in the numerator, the factor 12 and in the denominator, the factor 4 can be reduced by 4. We keep the four in our minds, and dividing 12 and 4 by this four, we write the answers next to these numbers, having previously crossed them out
Now you can multiply the resulting small factors. In this case, there are not many of them and you can multiply them in your mind:
Over time, you may find that when solving a particular problem, the expressions begin to “get fat”, so it is advisable to get used to fast calculations. What can be calculated in the mind must be calculated in the mind. What can be cut quickly should be cut quickly.
Example 4 Simplify Expression 
So the expression simplified to
Example 5 Simplify Expression 
We multiply numbers separately and letters separately:
So the expression simplified to mn.
Example 6 Simplify Expression 
Let's write this expression in more detail in order to clearly see where the numbers are and where the letters are:
Now we multiply the numbers separately and the letters separately. For convenience of calculations, the decimal fraction −6.4 and mixed number can be converted to ordinary fractions:
So the expression simplified to
The solution for this example can be written much shorter. It will look like this:
Example 7 Simplify Expression 
We multiply numbers separately and letters separately. For convenience of calculation, the mixed number and decimal fractions 0.1 and 0.6 can be converted to ordinary fractions:
So the expression simplified to abcd. If you skip the details, then this solution can be written much shorter:
Notice how the fraction has been reduced. New multipliers, which are obtained by reducing the previous multipliers, can also be reduced.
Now let's talk about what not to do. When simplifying expressions, it is strictly forbidden to multiply numbers and letters if the expression is a sum and not a product.
For example, if you want to simplify the expression 5a + 4b, then it cannot be written as follows:
This is equivalent to the fact that if we were asked to add two numbers, and we would multiply them instead of adding them.
When substituting any values of variables a and b expression 5a+4b turns into a simple numeric expression. Let's assume the variables a and b have the following meanings:
a = 2 , b = 3
Then the value of the expression will be 22
5a + 4b = 5 × 2 + 4 × 3 = 10 + 12 = 22
First, the multiplication is performed, and then the results are added. And if we tried to simplify this expression by multiplying numbers and letters, we would get the following:
5a + 4b = 5 × 4 × a × b = 20ab
20ab = 20 x 2 x 3 = 120
It turns out a completely different meaning of the expression. In the first case it turned out 22 , in the second case 120 . This means that the simplification of the expression 5a + 4b was performed incorrectly.
After simplifying the expression, its value should not change with the same values of the variables. If, when substituting any variable values into the original expression, one value is obtained, then after simplifying the expression, the same value should be obtained as before simplification.
With expression 5a + 4b actually nothing can be done. It doesn't get easier.
If the expression contains similar terms, then they can be added if our goal is to simplify the expression.
Example 8 Simplify Expression 0.3a−0.4a+a
0.3a − 0.4a + a = 0.3a + (−0.4a) + a = (0.3 + (−0.4) + 1)×a = 0.9a
or shorter: 0.3a - 0.4a + a = 0.9a
So the expression 0.3a−0.4a+a simplified to 0.9a
Example 9 Simplify Expression −7.5a − 2.5b + 4a
To simplify this expression, you can add like terms:
−7.5a − 2.5b + 4a = −7.5a + (−2.5b) + 4a = ((−7.5) + 4)×a + (−2.5b) = −3.5a + (−2.5b)
or shorter −7.5a − 2.5b + 4a = −3.5a + (−2.5b)
term (−2.5b) remained unchanged, since there was nothing to fold it with.
Example 10 Simplify Expression 
To simplify this expression, you can add like terms:
The coefficient was for the convenience of calculation.
So the expression simplified to
Example 11. Simplify Expression 
To simplify this expression, you can add like terms:
So the expression simplified to .
In this example, it would make more sense to add the first and last coefficient first. In this case, we would get a short solution. It would look like this:
Example 12. Simplify Expression 
To simplify this expression, you can add like terms:
So the expression simplified to 
.
The term remained unchanged, since there was nothing to add it to.
This solution can be written much shorter. It will look like this:
The short solution omits the steps of replacing subtraction with addition and a detailed record of how the fractions were reduced to a common denominator.
Another difference is that in the detailed solution, the answer looks like , but in short as . Actually, it's the same expression. The difference is that in the first case, subtraction is replaced by addition, because at the beginning, when we wrote down the solution in detail, we replaced subtraction with addition wherever possible, and this replacement was preserved for the answer.
Identities. Identical equal expressions
After we have simplified any expression, it becomes simpler and shorter. To check whether the expression is simplified correctly, it is enough to substitute any values of the variables first into the previous expression, which was required to be simplified, and then into the new one, which was simplified. If the value in both expressions is the same, then the expression is simplified correctly.
Consider the simplest example. Let it be required to simplify the expression 2a × 7b. To simplify this expression, you can separately multiply the numbers and letters:
2a × 7b = 2 × 7 × a × b = 14ab
Let's check if we simplified the expression correctly. To do this, substitute any values of the variables a and b first to the first expression, which needed to be simplified, and then to the second, which was simplified.
Let the values of the variables a , b will be as follows:
a = 4 , b = 5
Substitute them in the first expression 2a × 7b
Now let's substitute the same values of the variables into the expression that resulted from the simplification 2a×7b, namely in the expression 14ab
14ab = 14 x 4 x 5 = 280
We see that at a=4 and b=5 the value of the first expression 2a×7b and the value of the second expression 14ab equal
2a × 7b = 2 × 4 × 7 × 5 = 280
14ab = 14 x 4 x 5 = 280
The same will happen for any other values. For example, let a=1 and b=2
2a × 7b = 2 × 1 × 7 × 2 = 28
14ab = 14 x 1 x 2 = 28
Thus, for any values of the variables, the expressions 2a×7b and 14ab are equal to the same value. Such expressions are called identically equal.
We conclude that between the expressions 2a×7b and 14ab you can put an equal sign, since they are equal to the same value.
2a × 7b = 14ab
An equality is any expression that is joined by an equal sign (=).
And the equality of the form 2a×7b = 14ab called identity.
An identity is an equality that is true for any values of the variables.
Other examples of identities:
a + b = b + a
a(b+c) = ab + ac
a(bc) = (ab)c
Yes, the laws of mathematics that we studied are identities.
True numerical equalities are also identities. For example:
2 + 2 = 4
3 + 3 = 5 + 1
10 = 7 + 2 + 1
Deciding difficult task, in order to facilitate the calculation, the complex expression is replaced by a simpler expression that is identically equal to the previous one. Such a replacement is called identical transformation of the expression or simply expression conversion.
For example, we simplified the expression 2a × 7b, and get a simpler expression 14ab. This simplification can be called the identity transformation.
You can often find a task that says "prove that equality is identity" and then the equality to be proved is given. Usually this equality consists of two parts: the left and right parts of the equality. Our task is to perform identical transformations with one of the parts of the equality and get the other part. Or perform identical transformations with both parts of the equality and make sure that both parts of the equality contain the same expressions.
For example, let us prove that the equality 0.5a × 5b = 2.5ab is an identity.
Simplify the left side of this equality. To do this, multiply the numbers and letters separately:
0.5 × 5 × a × b = 2.5ab
2.5ab = 2.5ab
As a result of a small identity transformation, the left side of the equality became equal to the right side of the equality. So we have proved that the equality 0.5a × 5b = 2.5ab is an identity.
From identical transformations, we learned to add, subtract, multiply and divide numbers, reduce fractions, bring like terms, and also simplify some expressions.
But these are far from all identical transformations that exist in mathematics. There are many more identical transformations. We will see this again and again in the future.
Tasks for independent solution:
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