Equations with modules, methods of solutions. Part 1.

Before embarking on a direct study of the techniques for solving such equations, it is important to understand the essence of the module, its geometric meaning. It is in the understanding of the definition of the modulus and its geometric sense that the basic methods for solving such equations are laid. The so-called method of intervals when expanding modular brackets is so effective that using it it is possible to solve absolutely any equation or inequality with moduli. In this part, we will explore in detail two standard methods: the interval method and the equation replacement method with a set.

However, as we will see, these methods are always effective, but not always convenient and can lead to long and even inconvenient calculations, which will naturally require more time to solve them. Therefore, it is important to know those methods that greatly simplify the solution of certain structures of equations. Squaring both sides of the equation, a method for introducing a new variable, graphical method, solution of equations containing a modulus under the modulus sign. We'll look at these methods in the next part.

Determination of the modulus of a number. The geometric meaning of the module.

First of all, let's get acquainted with the geometric meaning of the module:

By the modulus of the number a (| a |) is the distance on the number line from the origin (point 0) to the point A (a).

Based on this definition, consider some examples:

|7| - this is the distance from 0 to point 7, of course it is equal to 7. → | 7 |=7

| -5 | is distance from 0 to point -5 and it is equal to: 5. → |-5| = 5

We all understand the distance cannot be negative! Therefore | x | ≥ 0 always!

Let's solve the equation: | x | = 4

This equation can be read as follows: the distance from point 0 to point x is 4. Yeah, it turns out that from 0 we can move both to the left and to the right, which means moving to the left at a distance equal to 4 we will find ourselves at the point: -4, and moving to the right we will find ourselves at point: 4. Indeed, | -4 | = 4 and | 4 | = 4.

Hence the answer is x = ± 4.

Upon closer examination of the previous equation, you will notice that: the distance to the right along the number line from 0 to the point is equal to the point itself, and the distance to the left from 0 to the number is equal to the opposite number! Realizing that there are positive numbers to the right of 0 and negative numbers to the left of 0, we formulate determining the modulus of a number: modulus (absolute value) of a number X(| x |) is the number itself X if x ≥0, and the number - X if x<0.

Here we need to find a set of points on the number line, the distance from 0 to which will be less than 3, let's imagine the number line, point 0 on it, go left and count one (-1), two (-2) and three (-3), stop. Further points will go that lie further than 3 or the distance to which from 0 is more than 3, now we go to the right: one, two, three, again stop. Now we select all our points and get the interval x: (- 3; 3).

It is important that you clearly see this, if it still does not work out, draw on paper and see that this illustration is completely understandable to you, do not be lazy and try to see the solutions to the following tasks in your mind:

| x | = 11, x =? | x | = -5, x =?

| x |<8, х-? |х| <-6, х-?

| x |> 2, x-? | x |> -3, x-?

| π-3 | =? | -x²-10 | =?

| √5-2 | =? | 2x-x²-3 | =?

| x² + 2 | =? | x² + 4 | = 0

| x² + 3x + 4 | =? | -x² + 9 | ≤0

Notice the weird quests in the second column? Indeed, the distance cannot be negative therefore: | x | = -5- has no solutions, of course it cannot be less than 0, therefore: | x |<-6 тоже не имеет решений, ну и естественно, что любое расстояние будет больше отрицательного числа, значит решением |x|>-3 are all numbers.

After you learn how to quickly see the pictures with solutions, read on.

We already know that the set real numbers$ R $ form rational and irrational numbers.

Rational numbers can always be represented as decimal fractions (finite or infinite periodic).

Irrational numbers are written as infinite but non-periodic decimal fractions.

The set of real numbers $ R $ also includes the elements $ - \ infty $ and $ + \ infty $, for which the inequalities $ - \ infty

Consider ways to represent real numbers.

Regular fractions

Ordinary fractions are written using two natural numbers and a horizontal fractional bar. The fractional slash actually replaces the division sign. The number under the line is the denominator of the fraction (divisor), the number above the line is the numerator (dividend).

Definition

A fraction is called correct if its numerator is less than the denominator. Conversely, a fraction is called incorrect if its numerator is greater than or equal to the denominator.

For ordinary fractions, there are simple, almost obvious, comparison rules ($ m $, $ n $, $ p $ are natural numbers):

1. of two fractions with the same denominator, the larger is the one with the larger numerator, that is, $ \ frac (m) (p)> \ frac (n) (p) $ for $ m> n $;
2. of two fractions with the same numerators, the larger is the one with the lower denominator, that is, $ \ frac (p) (m)> \ frac (p) (n) $ for $ m
3. a regular fraction is always less than one; improper fraction always greater than one; a fraction with the numerator equal to the denominator is equal to one;
4. any irregular fraction is greater than any correct one.

Decimal numbers

Decimal notation ( decimal) has the form: whole part, decimal point, fraction. Decimal notation the usual fraction can be obtained by dividing the "angle" of the numerator by the denominator. This can produce either a finite decimal fraction or an infinite periodic decimal fraction.

Definition

Fractional numbers are called decimal places. In this case, the first place after the decimal point is called the tenth place, the second - the hundredth place, the third - the thousandth place, etc.

Example 1

Determine the value of the decimal number 3.74. We get: $ 3.74 = 3 + \ frac (7) (10) + \ frac (4) (100) $.

The decimal number can be rounded. In this case, you should indicate the digit to which rounding is performed.

The rounding rule is as follows:

1. all digits to the right of this digit are replaced with zeros (if these digits are before the decimal point) or discarded (if these digits are after the decimal point);
2. if the first digit following this digit is less than 5, then the digit of this digit is not changed;
3. if the first digit following this digit is 5 or more, then the digit of this digit is increased by one.

Example 2

1. Let's round the number 17302 to thousands: 17000.
2. Let's round the number 17378 to hundreds: 17400.
3. Let's round the number 17378.45 to tens: 17380.
4. Let's round the number 378.91434 to hundredths: 378.91.
5. Let's round the number 378.91534 to hundredths: 378.92.

Convert a decimal number to a fraction.

Case 1

The decimal number is the final decimal fraction.

The conversion method is demonstrated by the following example.

Example 2

We have: $ 3.74 = 3 + \ frac (7) (10) + \ frac (4) (100) $.

We bring to common denominator and we get:

The fraction can be reduced: $ 3.74 = \ frac (374) (100) = \ frac (187) (50) $.

Case 2

A decimal number is an infinite periodic decimal fraction.

The conversion method is based on the fact that the periodic part of the periodic decimal fraction can be considered as the sum of the terms of an infinite decreasing geometric progression.

Example 4

$ 0, \ left (74 \ right) = \ frac (74) (100) + \ frac (74) (10000) + \ frac (74) (1000000) + \ ldots $. The first term of the progression is $ a = 0.74 $, the denominator of the progression is $ q = 0.01 $.

Example 5

$ 0.5 \ left (8 \ right) = \ frac (5) (10) + \ frac (8) (100) + \ frac (8) (1000) + \ frac (8) (10000) + \ ldots $ ... The first term of the progression is $ a = 0.08 $, the denominator of the progression is $ q = 0.1 $.

The sum of the terms of an infinite decreasing geometric progression is calculated by the formula $ s = \ frac (a) (1-q) $, where $ a $ is the first term, and $ q $ is the denominator of the progression $ \ left (0

Example 6

Let's convert the infinite periodic decimal fraction $ 0, \ left (72 \ right) $ into a regular one.

The first term of the progression is $ a = 0.72 $, the denominator of the progression is $ q = 0.01 $. We get: $ s = \ frac (a) (1-q) = \ frac (0.72) (1-0.01) = \ frac (0.72) (0.99) = \ frac (72) ( 99) = \ frac (8) (11) $. So $ 0, \ left (72 \ right) = \ frac (8) (11) $.

Example 7

Let's convert the infinite periodic decimal fraction $ 0.5 \ left (3 \ right) $ into a regular one.

The first term of the progression is $ a = 0.03 $, the denominator of the progression is $ q = 0.1 $. We get: $ s = \ frac (a) (1-q) = \ frac (0.03) (1-0.1) = \ frac (0.03) (0.9) = \ frac (3) ( 90) = \ frac (1) (30) $.

So $ 0.5 \ left (3 \ right) = \ frac (5) (10) + \ frac (1) (30) = \ frac (5 \ cdot 3) (10 \ cdot 3) + \ frac ( 1) (30) = \ frac (15) (30) + \ frac (1) (30) = \ frac (16) (30) = \ frac (8) (15) $.

Real numbers can be represented by points on the numerical axis.

In this case, we call the numerical axis an infinite straight line, on which the origin (point $ O $), positive direction (indicated by an arrow) and scale (for displaying values) are selected.

There is a one-to-one correspondence between all real numbers and all points of the numerical axis: each point corresponds singular conversely, each number has a single dot. Therefore, the set of real numbers is continuous and infinite, just as the number axis is continuous and infinite.

Some subsets of the set of real numbers are called numeric ranges. Elements of a numerical interval are numbers $ x \ in R $ satisfying a certain inequality. Let $ a \ in R $, $ b \ in R $ and $ a \ le b $. In this case, the types of gaps can be as follows:

1. Spacing $ \ left (a, \; b \ right) $. Moreover, $ a
2. Segment $ \ left $. Moreover, $ a \ le x \ le b $.
3. Semi-segments or half-intervals $ \ left $. Moreover, $ a \ le x
4. Infinite gaps, e.g. $ a

Also important is the type of gap called the neighborhood of the point. The neighborhood of a given point $ x_ (0) \ in R $ is an arbitrary interval $ \ left (a, \; b \ right) $, containing this point inside itself, that is, $ a 0 $ - with its radius.

The absolute value of the number

The absolute value (or modulus) of the real number $ x $ is the non-negative real number $ \ left | x \ right | $, determined by the formula: $ \ left | x \ right | = \ left \ (\ begin (array) (c) (\; \; x \; \; (\ rm for) \; \; x \ ge 0) \\ (-x \; \; (\ rm for) \; \; x

Geometrically, $ \ left | x \ right | $ means the distance between points $ x $ and 0 on the number axis.

Properties of absolute values:

1. it follows from the definition that $ \ left | x \ right | \ ge 0 $, $ \ left | x \ right | = \ left | -x \ right | $;
2. for the modulus of the sum and for the modulus of the difference of two numbers, the following inequalities hold: $ \ left | x + y \ right | \ le \ left | x \ right | + \ left | y \ right | $, $ \ left | xy \ right | \ le \ left | x \ right | + \ left | y \ right | $ as well as $ \ left | x + y \ right | \ ge \ left | x \ right | - \ left | y \ right | $, $ \ left | xy \ right | \ ge \ left | x \ right | - \ left | y \ right | $;
3. the modulus of the product and modulus of the quotient of two numbers satisfy the equalities $ \ left | x \ cdot y \ right | = \ left | x \ right | \ cdot \ left | y \ right | $ and $ \ left | \ frac (x) ( y) \ right | = \ frac (\ left | x \ right |) (\ left | y \ right |) $.

Based on the definition of the absolute value for an arbitrary number $ a> 0 $, one can also establish the equivalence of the following pairs of inequalities:

1. if $ \ left | x \ right |
2. if $ \ left | x \ right | \ le a $, then $ -a \ le x \ le a $;
3. if $ \ left | x \ right |> a $, then or $ xa $;
4. if $ \ left | x \ right | \ ge a $, then either $ x \ le -a $, or $ x \ ge a $.

Example 8

Solve the inequality $ \ left | 2 \ cdot x + 1 \ right |

This inequality is equivalent to the inequalities $ -7

From here we get: $ -8

The video lesson "The geometric meaning of the real number module" is a visual aid for a mathematics lesson on the relevant topic. The video tutorial examines in detail and visually geometric meaning module, after which the examples reveal how the module of a real number is, and the solution is accompanied by a figure. The material can be used during the explain stage. new topic as a separate part of the lesson or to provide clarity for the teacher's explanation. Both options contribute to an increase in the effectiveness of the math lesson, help the teacher to achieve the goals of the lesson.

This video tutorial contains constructions that clearly demonstrate the geometric meaning of the module. To make the demonstration more visual, these constructions are performed using animation effects. To educational material easier to remember, important theses are highlighted in color. The solution of examples is considered in detail, which, due to animation effects, is presented in a structured, sequential, understandable way. When compiling the video, tools were used that help to make the video tutorial an effective modern teaching tool.

The video begins by introducing the topic of the lesson. Construction is in progress on the screen - a ray is shown on which points a and b are marked, the distance between which is marked as ρ (a; b). It is recalled that the distance is measured on the coordinate ray by subtracting the smaller number from the larger one, that is, for this construction, the distance is equal to b-a for b> a and equal to a-b for a> b. Below is a construction where the marked point a lies to the right of b, that is, the corresponding numerical value is greater than b. Another case is noted below when the positions of points a and b coincide. In this case, the distance between the points is equal to zero ρ (a; b) = 0. Together, these cases are described by one formula ρ (a; b) = | a-b |.

Next, we consider the solution of problems in which knowledge about the geometric meaning of the module is applied. In the first example, you need to solve the equation | x-2 | = 3. It is noted that this is an analytical form of writing this equation, which we translate into geometric language to find a solution. Geometrically, this problem means that it is necessary to find points x for which the equality ρ (x; 2) = 3 will be true. On the coordinate line, this will mean the equidistance of points x from the point x = 2 at a distance 3. To demonstrate the solution on the coordinate line, a ray is drawn on which point 2 is marked. At a distance of 3 from the point x = 2, points -1 and 5 are marked. Obviously , that these marked points will be the solution to the equation.

To solve the equation | x + 3,2 | = 2, it is proposed to bring it first to the form | a-b | in order to solve the task on the coordinate line. After transformation, the equation takes the form | x - (- 3.2) | = 2. This means that the distance between the point -3.2 and the desired points will be equal to 2, that is, ρ (x; -3.2) = 2. The point -3.2 is marked on the coordinate line. From it at a distance of 2 points -1.2 and -5.2 are located. These points are marked on the coordinate line and indicated as the solution to the equation.

The solution to another equation | x | = 2.7 considers the case when the required points are located at a distance of 2.7 from point 0. The equation is rewritten as | x-0 | = 2.7. It is indicated that the distance to the desired points is determined as ρ (x; 0) = 2.7. The origin point 0 is marked on the coordinate line. Points -2.7 and 2.7 are located at a distance of 2.7 from point 0. These points are marked on the constructed straight line; they are the solutions of the equation.

To solve the following equation | x-√2 | = 0, no geometric interpretation is required, since if the modulus of the expression is zero, this means that this expression is equal to zero, that is, x-√2 = 0. It follows from the equation that x = √2.

The following example looks at solving equations that require transformation before solving. In the first equation | 2x-6 | = 8 before x there is a numerical coefficient 2. To get rid of the coefficient and translate the equation into geometric language ρ (x; a) = b, we put the common factor outside the parentheses, getting | 2 (x-3) | = 2 | x-3 |. After that, the right and left sides of the equation are canceled by 2. We get an equation of the form | x-3 | = 4. This analytic equation is translated into the geometric language ρ (x; 3) = 4. On the coordinate line, mark point 3. From this point, set aside points located at a distance of 4. The solution to the equation will be points -1 and 7, which are marked on the coordinate line. The second considered equation | 5-3x | = 6 also contains a numerical coefficient in front of the variable x. To solve the equation, the coefficient 3 is taken out of the parentheses. The equation becomes | -3 (x-5/3) | = 3 | x-5/3 |. The right and left sides of the equation can be canceled by 3. This gives an equation of the form | x-5/3 | = 2. We pass from the analytical form to the geometric interpretation ρ (x; 5/3) = 2. A drawing is constructed to the solution, which depicts the coordinate line. Point 5/3 is marked on this line. At a distance of 2 from point 5/3, points -1/3 and 11/3 are located. These points are the solutions of the equation.

The last equation considered | 4x + 1 | = -2. To solve this equation, no transformations and geometric representation are required. On the left-hand side of the equation, it obviously turns out not negative number and the right side contains the number -2. Therefore, this equation has no solutions.

The video lesson "The geometric meaning of the module of a real number" can be used in a traditional mathematics lesson at school. The material can be useful for a teacher exercising distance education... A detailed clear explanation of the solution of tasks that use the function of the module will help the student master the material, who is mastering the topic on his own.

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Goals:

Equipment: projector, screen, personal computer, multimedia presentation

During the classes

1. Organizational moment.

2. Actualization of students' knowledge.

2.1. Answer students' homework questions.

2.2. Solve the crossword puzzle (repetition of theoretical material) (Slide 2):

- Having solved the crossword puzzle, in the highlighted vertical column read the name of the topic of today's lesson. (Slides 3, 4)

3. Explanation of the new topic.

3.1. - Guys, you have already met the concept of a module, you used the designation | a| ... Previously, it was only about rational numbers... Now it is necessary to introduce the concept of a modulus for any real number.

Each real number corresponds to a single point on the number line, and, conversely, to each point on the number line corresponds to a single real number. All the basic properties of actions on rational numbers are preserved for real numbers.

The concept of the modulus of a real number is introduced. (Slide 5).

Definition. By the modulus of a non-negative real number x call this number itself: | x| = x; modulus of a negative real number X call the opposite number: | x| = – x .

– Write in notebooks the topic of the lesson, the definition of the module:

In practice, various module properties, For example. (Slide 6) :

Execute verbally No. 16.3 (a, b) - 16.5 (a, b) on the application of the definition, properties of the module. (Slide 7) .

3.4. For any real number X can be calculated | x| , i.e. we can talk about the function y = |x| .

Task 1. Build a graph and list the properties of the function y = |x| (Slides 8, 9).

One student draws a graph of a function on the blackboard

Fig 1.

Properties are enumerated by students. (Slide 10)

1) Domain of definition - (- ∞; + ∞).

2) y = 0 at x = 0; y> 0 for x< 0 и x > 0.

3) The function is continuous.

4) y naim = 0 for x = 0, y naib does not exist.

5) The function is limited at the bottom, not limited at the top.

6) The function decreases on the ray (- ∞; 0) and increases on the ray)