Module definitions. What is the modulus of a number in mathematics

Instructions

If a module is represented as a continuous function, then the value of its argument can be either positive or negative: |x| = x, x ≥ 0; |x| = - x, x

The modulus is zero, and the modulus of any positive number is . If the argument is negative, then after opening the brackets its sign changes from minus to plus. Based on this, the conclusion follows that the modules of opposites are equal: |-x| = |x| = x.

The modulus of a complex number is found by the formula: |a| = √b ² + c ², and |a + b| ≤ |a| + |b|. If the argument contains a positive number as a multiplier, then it can be taken out of the bracket sign, for example: |4*b| = 4*|b|.

If the argument is presented as a complex number, then for convenience of calculations the order of the terms of the expression enclosed in rectangular brackets is allowed: |2-3| = |3-2| = 3-2 = 1 because (2-3) is less than zero.

The argument raised to a power is simultaneously under the sign of a root of the same order - it is solved using: √a² = |a| = ±a.

If you have a task in which the condition for expanding the module brackets is not specified, then there is no need to get rid of them - this will be the end result. And if you need to open them, then you must indicate the ± sign. For example, you need to find the value of the expression √(2 * (4-b))². His solution looks like this: √(2 * (4-b))² = |2 * (4-b)| = 2 * |4-b|. Since the sign of expression 4-b is unknown, it must be left in parentheses. If you add an additional condition, for example, |4-b| >

The modulus of zero is equal to zero, and the modulus of any positive number is equal to itself. If the argument is negative, then after opening the brackets its sign changes from minus to plus. Based on this, the conclusion follows that the modules of opposite numbers are equal: |-x| = |x| = x.

The modulus of a complex number is found by the formula: |a| = √b ² + c ², and |a + b| ≤ |a| + |b|. If the argument contains a positive integer as a factor, then it can be taken out of the bracket sign, for example: |4*b| = 4*|b|.

The modulus cannot be negative, so any negative number is converted to positive: |-x| = x, |-2| = 2, |-1/7| = 1/7, |-2.5| = 2.5.

If the argument is presented in the form of a complex number, then for the convenience of calculations it is allowed to change the order of the terms of the expression enclosed in rectangular brackets: |2-3| = |3-2| = 3-2 = 1 because (2-3) is less than zero.

If you have a task in which the condition for expanding the module brackets is not specified, then there is no need to get rid of them - this will be the end result. And if you need to open them, then you must indicate the ± sign. For example, you need to find the value of the expression √(2 * (4-b))². His solution looks like this: √(2 * (4-b))² = |2 * (4-b)| = 2 * |4-b|. Since the sign of expression 4-b is unknown, it must be left in parentheses. If you add an additional condition, for example, |4-b| > 0, then the result will be 2 * |4-b| = 2 *(4 - b). The unknown element can also be set to a specific number, which should be taken into account because it will influence the sign of the expression.

Modulus of numbers this number itself is called if it is non-negative, or the same number with the opposite sign if it is negative.

For example, the modulus of the number 5 is 5, and the modulus of the number –5 is also 5.

That is, the modulus of a number is understood as the absolute value, the absolute value of this number without taking into account its sign.

Denoted as follows: |5|, | X|, |A| etc.

Rule:

Explanation:

|5| = 5
It reads like this: the modulus of the number 5 is 5.

|–5| = –(–5) = 5
It reads like this: the modulus of the number –5 is 5.

|0| = 0
It reads like this: the modulus of zero is zero.

Module properties:

Geometric meaning of the module.

The modulus of a number is the distance from zero to that number.

For example, let's take the number 5 again. The distance from 0 to 5 is the same as from 0 to –5 (Fig. 1). And when it is important for us to know only the length of the segment, then the sign has not only meaning, but also meaning. However, this is not entirely true: we measure distance only with positive numbers - or non-negative numbers. Let the division price of our scale be 1 cm. Then the length of the segment from zero to 5 is 5 cm, from zero to –5 is also 5 cm.

In practice, the distance is often measured not only from zero - the reference point can be any number (Fig. 2). But this does not change the essence. Notation of the form |a – b| expresses the distance between points A And b on the number line.

Example 1. Solve the equation | X – 1| = 3.

Solution .

The meaning of the equation is that the distance between points X and 1 is equal to 3 (Fig. 2). Therefore, from point 1 we count three divisions to the left and three divisions to the right - and we clearly see both values X:
X 1 = –2, X 2 = 4.

We can calculate it.

│X – 1 = 3
│X – 1 = –3

│X = 3 + 1
│X = –3 + 1

│X = 4
│ X = –2.

Answer : X 1 = –2; X 2 = 4.

Example 2. Find expression module:

Solution .

First, let's find out whether the expression is positive or negative. To do this, we transform the expression so that it consists of homogeneous numbers. Let's not look for the root of 5 - it's quite difficult. Let's do it simpler: let's raise 3 and 10 to the root. Then compare the magnitude of the numbers that make up the difference:

3 = √9. Therefore, 3√5 = √9 √5 = √45

10 = √100.

We see that the first number is less than the second. This means that the expression is negative, that is, its answer is less than zero:

3√5 – 10 < 0.

But according to the rule, the modulus of a negative number is the same number with the opposite sign. We have a negative expression. Therefore, it is necessary to change its sign to the opposite one. The opposite of 3√5 – 10 is –(3√5 – 10). Let's open the brackets in it and get the answer:

–(3√5 – 10) = –3√5 + 10 = 10 – 3√5.

Answer .

The module is one of those things that everyone seems to have heard about, but in reality no one really understands. Therefore, today there will be a big lesson dedicated to solving equations with modules.

I’ll say right away: the lesson will not be difficult. And in general, modules are a relatively simple topic. “Yes, of course, it’s not complicated! It blows my mind!” - many students will say, but all these brain breaks occur due to the fact that most people do not have knowledge in their heads, but some kind of crap. And the goal of this lesson is to turn crap into knowledge. :)

A little theory

So, let's go. Let's start with the most important thing: what is a module? Let me remind you that the modulus of a number is simply the same number, but taken without the minus sign. That is, for example, $\left| -5 \right|=5$. Or $\left| -129.5 \right|=$129.5.

Is it that simple? Yes, simple. What then is the absolute value of a positive number? It’s even simpler here: the modulus of a positive number is equal to this number itself: $\left| 5 \right|=5$; $\left| 129.5 \right|=$129.5, etc.

It turns out a curious thing: different numbers can have the same module. For example: $\left| -5 \right|=\left| 5 \right|=5$; $\left| -129.5 \right|=\left| 129.5\right|=$129.5. It is easy to see what kind of numbers these are, whose modules are the same: these numbers are opposite. Thus, we note for ourselves that the modules of opposite numbers are equal:

\[\left| -a \right|=\left| a\right|\]

Another important fact: modulus is never negative. Whatever number we take - be it positive or negative - its modulus always turns out to be positive (or, in extreme cases, zero). This is why the modulus is often called the absolute value of a number.

In addition, if we combine the definition of the modulus for a positive and negative number, we obtain a global definition of the modulus for all numbers. Namely: the modulus of a number is equal to the number itself if the number is positive (or zero), or equal to the opposite number if the number is negative. You can write this as a formula:

There is also a modulus of zero, but it is always equal to zero. In addition, zero is the only number that does not have an opposite.

Thus, if we consider the function $y=\left| x \right|$ and try to draw its graph, you will get something like this:

Modulus graph and example of solving the equation

From this picture it is immediately clear that $\left| -m \right|=\left| m \right|$, and the modulus graph never falls below the x-axis. But that’s not all: the red line marks the straight line $y=a$, which, for positive $a$, gives us two roots at once: $((x)_(1))$ and $((x)_(2)) $, but we'll talk about that later. :)

In addition to the purely algebraic definition, there is a geometric one. Let's say there are two points on the number line: $((x)_(1))$ and $((x)_(2))$. In this case, the expression $\left| ((x)_(1))-((x)_(2)) \right|$ is simply the distance between the specified points. Or, if you prefer, the length of the segment connecting these points:

This definition also implies that the modulus is always non-negative. But enough definitions and theory - let's move on to real equations. :)

Basic formula

Okay, we've sorted out the definition. But that didn’t make it any easier. How to solve equations containing this very module?

Calm, just calm. Let's start with the simplest things. Consider something like this:

\[\left| x\right|=3\]

So the modulus of $x$ is 3. What could $x$ be equal to? Well, judging by the definition, we are quite happy with $x=3$. Really:

\[\left| 3\right|=3\]

Are there other numbers? Cap seems to be hinting that there is. For example, $x=-3$ is also $\left| -3 \right|=3$, i.e. the required equality is satisfied.

So maybe if we search and think, we will find more numbers? But let's face it: there are no more numbers. Equation $\left| x \right|=3$ has only two roots: $x=3$ and $x=-3$.

Now let's complicate the task a little. Let the function $f\left(x \right)$ hang out under the modulus sign instead of the variable $x$, and on the right instead of the triple we put an arbitrary number $a$. We get the equation:

\[\left| f\left(x \right) \right|=a\]

So how can we solve this? Let me remind you: $f\left(x \right)$ is an arbitrary function, $a$ is any number. Those. Anything at all! For example:

\[\left| 2x+1 \right|=5\]

\[\left| 10x-5 \right|=-65\]

Let's pay attention to the second equation. You can immediately say about him: he has no roots. Why? Everything is correct: because it requires that the modulus be equal to a negative number, which never happens, since we already know that the modulus is always a positive number or, in extreme cases, zero.

But with the first equation everything is more fun. There are two options: either there is a positive expression under the modulus sign, and then $\left| 2x+1 \right|=2x+1$, or this expression is still negative, and then $\left| 2x+1 \right|=-\left(2x+1 \right)=-2x-1$. In the first case, our equation will be rewritten as follows:

\[\left| 2x+1 \right|=5\Rightarrow 2x+1=5\]

And suddenly it turns out that the submodular expression $2x+1$ is really positive - it is equal to the number 5. That is we can safely solve this equation - the resulting root will be a piece of the answer:

Those who are especially distrustful can try to substitute the found root into the original equation and make sure that there really is a positive number under the modulus.

Now let's look at the case of a negative submodular expression:

\[\left\( \begin(align)& \left| 2x+1 \right|=5 \\& 2x+1 \lt 0 \\\end(align) \right.\Rightarrow -2x-1=5 \Rightarrow 2x+1=-5\]

Oops! Again, everything is clear: we assumed that $2x+1 \lt 0$, and as a result we got that $2x+1=-5$ - indeed, this expression is less than zero. We solve the resulting equation, while already knowing for sure that the found root will suit us:

In total, we again received two answers: $x=2$ and $x=3$. Yes, the amount of calculations turned out to be a little larger than in the very simple equation $\left| x \right|=3$, but nothing fundamentally has changed. So maybe there is some kind of universal algorithm?

Yes, such an algorithm exists. And now we will analyze it.

Getting rid of the modulus sign

Let us be given the equation $\left| f\left(x \right) \right|=a$, and $a\ge 0$ (otherwise, as we already know, there are no roots). Then you can get rid of the modulus sign using the following rule:

\[\left| f\left(x \right) \right|=a\Rightarrow f\left(x \right)=\pm a\]

Thus, our equation with a modulus splits into two, but without a modulus. That's all the technology is! Let's try to solve a couple of equations. Let's start with this

\[\left| 5x+4 \right|=10\Rightarrow 5x+4=\pm 10\]

Let’s consider separately when there is a ten plus on the right, and separately when there is a minus. We have:

\[\begin(align)& 5x+4=10\Rightarrow 5x=6\Rightarrow x=\frac(6)(5)=1,2; \\& 5x+4=-10\Rightarrow 5x=-14\Rightarrow x=-\frac(14)(5)=-2.8. \\\end(align)\]

That's all! We got two roots: $x=1.2$ and $x=-2.8$. The entire solution took literally two lines.

Ok, no question, let's look at something a little more serious:

\[\left| 7-5x\right|=13\]

Again we open the module with plus and minus:

\[\begin(align)& 7-5x=13\Rightarrow -5x=6\Rightarrow x=-\frac(6)(5)=-1,2; \\& 7-5x=-13\Rightarrow -5x=-20\Rightarrow x=4. \\\end(align)\]

A couple of lines again - and the answer is ready! As I said, there is nothing complicated about modules. You just need to remember a few rules. Therefore, we move on and begin with truly more complex tasks.

The case of a right-hand side variable

Now consider this equation:

\[\left| 3x-2 \right|=2x\]

This equation is fundamentally different from all previous ones. How? And the fact that to the right of the equal sign is the expression $2x$ - and we cannot know in advance whether it is positive or negative.

What to do in this case? First, we must understand once and for all that if the right side of the equation turns out to be negative, then the equation will have no roots- we already know that the module cannot be equal to a negative number.

And secondly, if the right part is still positive (or equal to zero), then you can act in exactly the same way as before: simply open the module separately with a plus sign and separately with a minus sign.

Thus, we formulate a rule for arbitrary functions $f\left(x \right)$ and $g\left(x \right)$ :

\[\left| f\left(x \right) \right|=g\left(x \right)\Rightarrow \left\( \begin(align)& f\left(x \right)=\pm g\left(x \right ), \\& g\left(x \right)\ge 0. \\\end(align) \right.\]

In relation to our equation we get:

\[\left| 3x-2 \right|=2x\Rightarrow \left\( \begin(align)& 3x-2=\pm 2x, \\& 2x\ge 0. \\\end(align) \right.\]

Well, we will somehow cope with the requirement $2x\ge 0$. In the end, we can stupidly substitute the roots that we get from the first equation and check whether the inequality holds or not.

So let’s solve the equation itself:

\[\begin(align)& 3x-2=2\Rightarrow 3x=4\Rightarrow x=\frac(4)(3); \\& 3x-2=-2\Rightarrow 3x=0\Rightarrow x=0. \\\end(align)\]

Well, which of these two roots satisfies the requirement $2x\ge 0$? Yes both! Therefore, the answer will be two numbers: $x=(4)/(3)\;$ and $x=0$. That's the solution. :)

I suspect that some of the students are already starting to get bored? Well, let's look at an even more complex equation:

\[\left| ((x)^(3))-3((x)^(2))+x \right|=x-((x)^(3))\]

Although it looks evil, in fact it is still the same equation of the form “modulus equals function”:

\[\left| f\left(x \right) \right|=g\left(x \right)\]

And it is solved in exactly the same way:

\[\left| ((x)^(3))-3((x)^(2))+x \right|=x-((x)^(3))\Rightarrow \left\( \begin(align)& ( (x)^(3))-3((x)^(2))+x=\pm \left(x-((x)^(3)) \right), \\& x-((x )^(3))\ge 0. \\\end(align) \right.\]

We will deal with inequality later - it is somehow too evil (in fact, it is simple, but we will not solve it). For now, it’s better to deal with the resulting equations. Let's consider the first case - this is when the module is expanded with a plus sign:

\[((x)^(3))-3((x)^(2))+x=x-((x)^(3))\]

Well, it’s a no brainer that you need to collect everything from the left, bring similar ones and see what happens. And this is what happens:

\[\begin(align)& ((x)^(3))-3((x)^(2))+x=x-((x)^(3)); \\& 2((x)^(3))-3((x)^(2))=0; \\\end(align)\]

We take the common factor $((x)^(2))$ out of brackets and get a very simple equation:

\[((x)^(2))\left(2x-3 \right)=0\Rightarrow \left[ \begin(align)& ((x)^(2))=0 \\& 2x-3 =0 \\\end(align) \right.\]

\[((x)_(1))=0;\quad ((x)_(2))=\frac(3)(2)=1.5.\]

Here we took advantage of an important property of the product, for the sake of which we factored the original polynomial: the product is equal to zero when at least one of the factors is equal to zero.

Now let’s deal with the second equation in exactly the same way, which is obtained by expanding the module with a minus sign:

\[\begin(align)& ((x)^(3))-3((x)^(2))+x=-\left(x-((x)^(3)) \right); \\& ((x)^(3))-3((x)^(2))+x=-x+((x)^(3)); \\& -3((x)^(2))+2x=0; \\& x\left(-3x+2 \right)=0. \\\end(align)\]

Again the same thing: the product is equal to zero when at least one of the factors is equal to zero. We have:

\[\left[ \begin(align)& x=0 \\& -3x+2=0 \\\end(align) \right.\]

Well, we got three roots: $x=0$, $x=1.5$ and $x=(2)/(3)\;$. Well, which of this set will go into the final answer? To do this, remember that we have an additional constraint in the form of inequality:

How to take this requirement into account? Let’s just substitute the found roots and check whether the inequality holds for these $x$ or not. We have:

\[\begin(align)& x=0\Rightarrow x-((x)^(3))=0-0=0\ge 0; \\& x=1.5\Rightarrow x-((x)^(3))=1.5-((1.5)^(3)) \lt 0; \\& x=\frac(2)(3)\Rightarrow x-((x)^(3))=\frac(2)(3)-\frac(8)(27)=\frac(10) (27)\ge 0; \\\end(align)\]

Thus, the root $x=1.5$ does not suit us. And in response there will be only two roots:

\[((x)_(1))=0;\quad ((x)_(2))=\frac(2)(3).\]

As you can see, even in this case there was nothing complicated - equations with modules are always solved using an algorithm. You just need to have a good understanding of polynomials and inequalities. Therefore, we move on to more complex tasks - there will already be not one, but two modules.

Equations with two modules

Until now, we have studied only the simplest equations - there was one module and something else. We sent this “something else” to another part of the inequality, away from the module, so that in the end everything would be reduced to an equation of the form $\left| f\left(x \right) \right|=g\left(x \right)$ or even simpler $\left| f\left(x \right) \right|=a$.

But kindergarten is over - it's time to consider something more serious. Let's start with equations like this:

\[\left| f\left(x \right) \right|=\left| g\left(x \right) \right|\]

This is an equation of the form “modulus equals modulus”. A fundamentally important point is the absence of other terms and factors: only one module on the left, one more module on the right - and nothing more.

Someone will now think that such equations are more difficult to solve than what we have studied so far. But no: these equations are even easier to solve. Here's the formula:

\[\left| f\left(x \right) \right|=\left| g\left(x \right) \right|\Rightarrow f\left(x \right)=\pm g\left(x \right)\]

All! We simply equate submodular expressions by putting a plus or minus sign in front of one of them. And then we solve the resulting two equations - and the roots are ready! No additional restrictions, no inequalities, etc. Everything is very simple.

Let's try to solve this problem:

\[\left| 2x+3 \right|=\left| 2x-7 \right|\]

Elementary Watson! Expanding the modules:

\[\left| 2x+3 \right|=\left| 2x-7 \right|\Rightarrow 2x+3=\pm \left(2x-7 \right)\]

Let's consider each case separately:

\[\begin(align)& 2x+3=2x-7\Rightarrow 3=-7\Rightarrow \emptyset ; \\& 2x+3=-\left(2x-7 \right)\Rightarrow 2x+3=-2x+7. \\\end(align)\]

The first equation has no roots. Because when is $3=-7$? At what values ​​of $x$? “What the hell is $x$? Are you stoned? There’s no $x$ there at all,” you say. And you'll be right. We have obtained an equality that does not depend on the variable $x$, and at the same time the equality itself is incorrect. That's why there are no roots. :)

With the second equation, everything is a little more interesting, but also very, very simple:

As you can see, everything was solved literally in a couple of lines - we didn’t expect anything else from a linear equation. :)

As a result, the final answer is: $x=1$.

So how? Difficult? Of course not. Let's try something else:

\[\left| x-1 \right|=\left| ((x)^(2))-3x+2 \right|\]

Again we have an equation of the form $\left| f\left(x \right) \right|=\left| g\left(x \right) \right|$. Therefore, we immediately rewrite it, revealing the modulus sign:

\[((x)^(2))-3x+2=\pm \left(x-1 \right)\]

Perhaps someone will now ask: “Hey, what nonsense? Why does “plus-minus” appear on the right expression and not on the left?” Calm down, I’ll explain everything now. Indeed, in a good way we should have rewritten our equation as follows:

Then you need to open the brackets, move all the terms to one side of the equal sign (since the equation, obviously, will be square in both cases), and then find the roots. But you must admit: when “plus-minus” appears before three terms (especially when one of these terms is a quadratic expression), it somehow looks more complicated than the situation when “plus-minus” appears before only two terms.

But nothing prevents us from rewriting the original equation as follows:

\[\left| x-1 \right|=\left| ((x)^(2))-3x+2 \right|\Rightarrow \left| ((x)^(2))-3x+2 \right|=\left| x-1 \right|\]

What happened? Nothing special: they just swapped the left and right sides. A little thing that will ultimately make our life a little easier. :)

In general, we solve this equation, considering options with a plus and a minus:

\[\begin(align)& ((x)^(2))-3x+2=x-1\Rightarrow ((x)^(2))-4x+3=0; \\& ((x)^(2))-3x+2=-\left(x-1 \right)\Rightarrow ((x)^(2))-2x+1=0. \\\end(align)\]

The first equation has roots $x=3$ and $x=1$. The second is generally an exact square:

\[((x)^(2))-2x+1=((\left(x-1 \right))^(2))\]

Therefore, it has only one root: $x=1$. But we have already obtained this root earlier. Thus, only two numbers will go into the final answer:

\[((x)_(1))=3;\quad ((x)_(2))=1.\]

Mission Complete! You can take a pie from the shelf and eat it. There are 2 of them, yours is the middle one. :)

Important Note. The presence of identical roots for different variants of expansion of the module means that the original polynomials are factorized, and among these factors there will definitely be a common one. Really:

\[\begin(align)& \left| x-1 \right|=\left| ((x)^(2))-3x+2 \right|; \\& \left| x-1 \right|=\left| \left(x-1 \right)\left(x-2 \right) \right|. \\\end(align)\]

One of the module properties: $\left| a\cdot b \right|=\left| a \right|\cdot \left| b \right|$ (i.e. the modulus of the product is equal to the product of the moduli), so the original equation can be rewritten as follows:

\[\left| x-1 \right|=\left| x-1 \right|\cdot \left| x-2 \right|\]

As you can see, we really have a common factor. Now, if you collect all the modules on one side, you can take this factor out of the bracket:

\[\begin(align)& \left| x-1 \right|=\left| x-1 \right|\cdot \left| x-2 \right|; \\& \left| x-1 \right|-\left| x-1 \right|\cdot \left| x-2 \right|=0; \\& \left| x-1 \right|\cdot \left(1-\left| x-2 \right| \right)=0. \\\end(align)\]

Well, now remember that the product is equal to zero when at least one of the factors is equal to zero:

\[\left[ \begin(align)& \left| x-1 \right|=0, \\& \left| x-2 \right|=1. \\\end(align) \right.\]

Thus, the original equation with two modules has been reduced to the two simplest equations that we talked about at the very beginning of the lesson. Such equations can be solved literally in a couple of lines. :)

This remark may seem unnecessarily complex and inapplicable in practice. However, in reality, you may encounter much more complex problems than those we are looking at today. In them, modules can be combined with polynomials, arithmetic roots, logarithms, etc. And in such situations, the ability to lower the overall degree of the equation by taking something out of brackets can be very, very useful. :)

Now I would like to analyze another equation, which at first glance may seem crazy. Many students get stuck on it, even those who think they have a good understanding of the modules.

However, this equation is even easier to solve than what we looked at earlier. And if you understand why, you'll get another trick for quickly solving equations with moduli.

So the equation is:

\[\left| x-((x)^(3)) \right|+\left| ((x)^(2))+x-2 \right|=0\]

No, this is not a typo: there is a plus between the modules. And we need to find at what $x$ the sum of two modules is equal to zero. :)

What's the problem anyway? But the problem is that each module is a positive number, or, in extreme cases, zero. What happens if you add two positive numbers? Obviously a positive number again:

\[\begin(align)& 5+7=12 \gt 0; \\& 0.004+0.0001=0.0041 \gt 0; \\& 5+0=5 \gt 0. \\\end(align)\]

The last line might give you an idea: the only time the sum of the modules is zero is if each module is zero:

\[\left| x-((x)^(3)) \right|+\left| ((x)^(2))+x-2 \right|=0\Rightarrow \left\( \begin(align)& \left| x-((x)^(3)) \right|=0, \\& \left| ((x)^(2))+x-2 \right|=0. \\\end(align) \right.\]

And when is the module equal to zero? Only in one case - when the submodular expression is equal to zero:

\[((x)^(2))+x-2=0\Rightarrow \left(x+2 \right)\left(x-1 \right)=0\Rightarrow \left[ \begin(align)& x=-2 \\& x=1 \\\end(align) \right.\]

Thus, we have three points at which the first module is reset to zero: 0, 1 and −1; as well as two points at which the second module is reset to zero: −2 and 1. However, we need both modules to be reset to zero at the same time, so among the found numbers we need to choose those that are included in both sets. Obviously, there is only one such number: $x=1$ - this will be the final answer.

Cleavage method

Well, we've already covered a bunch of problems and learned a lot of techniques. Do you think that's all? But no! Now we will look at the final technique - and at the same time the most important. We will talk about splitting equations with modulus. What will we even talk about? Let's go back a little and look at some simple equation. For example this:

\[\left| 3x-5 \right|=5-3x\]

In principle, we already know how to solve such an equation, because it is a standard construction of the form $\left| f\left(x \right) \right|=g\left(x \right)$. But let's try to look at this equation from a slightly different angle. More precisely, consider the expression under the modulus sign. Let me remind you that the modulus of any number can be equal to the number itself, or it can be opposite to this number:

\[\left| a \right|=\left\( \begin(align)& a,\quad a\ge 0, \\& -a,\quad a \lt 0. \\\end(align) \right.\]

Actually, this ambiguity is the whole problem: since the number under the modulus changes (it depends on the variable), it is not clear to us whether it is positive or negative.

But what if you initially require that this number be positive? For example, let's require that $3x-5 \gt 0$ - in this case we are guaranteed to get a positive number under the modulus sign, and we can completely get rid of this very modulus:

Thus, our equation will turn into a linear one, which can be easily solved:

True, all these thoughts make sense only under the condition $3x-5 \gt 0$ - we ourselves introduced this requirement in order to unambiguously reveal the module. Therefore, let's substitute the found $x=\frac(5)(3)$ into this condition and check:

It turns out that for the specified value of $x$ our requirement is not met, because the expression turned out to be equal to zero, and we need it to be strictly greater than zero. Sad. :(

But it's okay! After all, there is another option $3x-5 \lt 0$. Moreover: there is also the case $3x-5=0$ - this also needs to be considered, otherwise the solution will be incomplete. So, consider the case $3x-5 \lt 0$:

Obviously, the module will open with a minus sign. But then a strange situation arises: both on the left and on the right in the original equation the same expression will stick out:

I wonder at what $x$ the expression $5-3x$ will be equal to the expression $5-3x$? Even Captain Obviousness would choke on his saliva from such equations, but we know: this equation is an identity, i.e. it is true for any value of the variable!

This means that any $x$ will suit us. However, we have a limitation:

In other words, the answer will not be a single number, but a whole interval:

Finally, there is one more case left to consider: $3x-5=0$. Everything is simple here: under the modulus there will be zero, and the modulus of zero is also equal to zero (this follows directly from the definition):

But then the original equation $\left| 3x-5 \right|=5-3x$ will be rewritten as follows:

We already obtained this root above when we considered the case of $3x-5 \gt 0$. Moreover, this root is a solution to the equation $3x-5=0$ - this is the limitation that we ourselves introduced to reset the module. :)

Thus, in addition to the interval, we will also be satisfied with the number lying at the very end of this interval:

Total final answer: $x\in \left(-\infty ;\frac(5)(3) \right]$ It’s not very common to see such crap in the answer to a fairly simple (essentially linear) equation with modulus , really? Well, get used to it: the difficulty of the module is that the answers in such equations can be completely unpredictable.

Something else is much more important: we have just analyzed a universal algorithm for solving an equation with a modulus! And this algorithm consists of the following steps:

1. Equate each modulus in the equation to zero. We get several equations;
2. Solve all these equations and mark the roots on the number line. As a result, the straight line will be divided into several intervals, at each of which all modules are uniquely revealed;
3. Solve the original equation for each interval and combine your answers.

That's all! There is only one question left: what to do with the roots obtained in step 1? Let's say we have two roots: $x=1$ and $x=5$. They will split the number line into 3 pieces:

So what are the intervals? It is clear that there are three of them:

1. The leftmost one: $x \lt 1$ — the unit itself is not included in the interval;
2. Central: $1\le x \lt 5$ - here one is included in the interval, but five is not included;
3. Rightmost: $x\ge 5$ - five is only included here!

I think you already understand the pattern. Each interval includes the left end and does not include the right.

At first glance, such an entry may seem inconvenient, illogical and generally some kind of crazy. But believe me: after a little practice, you will find that this approach is the most reliable and does not interfere with unambiguously opening the modules. It’s better to use such a scheme than to think every time: give the left/right end to the current interval or “throw” it into the next one.

This concludes the lesson. Download problems to solve on your own, practice, compare with the answers - and see you in the next lesson, which will be devoted to inequalities with moduli. :)

First we define the expression sign under the module sign, and then we expand the module:

• if the value of the expression is greater than zero, then we simply remove it from under the modulus sign,
• if the expression is less than zero, then we remove it from under the modulus sign, changing the sign, as we did earlier in the examples.

Well, shall we try? Let's evaluate:

(Forgot, Repeat.)

If so, what sign does it have? Well, of course, !

And, therefore, we expand the sign of the module by changing the sign of the expression:

Got it? Then try it yourself:

Answers:

What other properties does the module have?

If we need to multiply numbers inside the modulus sign, we can easily multiply the moduli of these numbers!!!

In mathematical terms, The modulus of the product of numbers is equal to the product of the moduli of these numbers.

For example:

What if we need to divide two numbers (expressions) under the modulus sign?

Yes, the same as with multiplication! Let's break it down into two separate numbers (expressions) under the modulus sign:

provided that (since you cannot divide by zero).

It is worth remembering one more property of the module:

The modulus of the sum of numbers is always less than or equal to the sum of the moduli of these numbers:

Why is that? Everything is very simple!

As we remember, the modulus is always positive. But under the modulus sign there can be any number: both positive and negative. Let's assume that the numbers and are both positive. Then the left expression will be equal to the right expression.

Let's look at an example:

If under the modulus sign one number is negative and the other is positive, the left expression will always be less than the right one:

Everything seems clear with this property, let’s look at a couple more useful properties of the module.

What if we have this expression:

What can we do with this expression? The value of x is unknown to us, but we already know what, which means.

The number is greater than zero, which means you can simply write:

So we come to another property, which in general can be represented as follows:

What does this expression equal:

So, we need to define the sign under the modulus. Is it necessary to define a sign here?

Of course not, if you remember that any number squared is always greater than zero! If you don't remember, see the topic. So what happens? Here's what:

Great, right? Quite convenient. And now a specific example to consolidate:

Well, why the doubts? Let's act boldly!

Have you figured it all out? Then go ahead and practice with examples!

1. Find the value of the expression if.

2. Which numbers have the same modulus?

3. Find the meaning of the expressions:

If not everything is clear yet and there are difficulties in solutions, then let’s figure it out:

Solution 1:

So, let’s substitute the values ​​and into the expression

Solution 2:

As we remember, opposite numbers are equal in modulus. This means that the modulus value is equal to two numbers: and.

Solution 3:

A)
b)
V)
G)

Did you catch everything? Then it's time to move on to something more complex!

Let's try to simplify the expression

Solution:

So, we remember that the modulus value cannot be less than zero. If the modulus sign has a positive number, then we can simply discard the sign: the modulus of the number will be equal to this number.

But if there is a negative number under the modulus sign, then the modulus value is equal to the opposite number (that is, the number taken with the “-” sign).

In order to find the modulus of any expression, you first need to find out whether it takes a positive or negative value.

It turns out that the value of the first expression under the module.

Therefore, the expression under the modulus sign is negative. The second expression under the modulus sign is always positive, since we are adding two positive numbers.

So, the value of the first expression under the modulus sign is negative, the second is positive:

This means that when expanding the modulus sign of the first expression, we must take this expression with the “-” sign. Like this:

In the second case, we simply discard the modulus sign:

Let's simplify this expression in its entirety:

Module of number and its properties (rigorous definitions and proofs)

Definition:

The modulus (absolute value) of a number is the number itself, if, and the number, if:

For example:

Example:

Simplify the expression.

Solution:

Basic properties of the module

For all:

Example:

Prove property No. 5.

Proof:

Let us assume that there are such that

Let's square the left and right sides of the inequality (this can be done, since both sides of the inequality are always non-negative):

and this contradicts the definition of a module.

Consequently, such people do not exist, which means that the inequality holds for all

Examples for independent solutions:

1) Prove property No. 6.

2) Simplify the expression.

Answers:

1) Let's use property No. 3: , and since, then

To simplify, you need to expand the modules. And to expand modules, you need to find out whether the expressions under the module are positive or negative?

a.

Let's compare the numbers and and:

b.

Now let's compare:

We add up the values ​​of the modules:

Module properties:

1. The absolute value of a number. Briefly about the main thing.
2. The modulus (absolute value) of a number is the number itself, if, and the number, if:
3. The modulus of a number is a non-negative number: ;
4. The modules of opposite numbers are equal: ;
5. The modulus of the product of two (or more) numbers is equal to the product of their moduli: ;
6. The modulus of the quotient of two numbers is equal to the quotient of their moduli: ;

The modulus of the sum of numbers is always less than or equal to the sum of the moduli of these numbers: ;

A constant positive multiplier can be taken out of the modulus sign: at;

Module of number is a new concept in mathematics. Let's take a closer look at what a number module is and how to work with it?

Let's look at an example:

We left the house to go to the store. We walked 300 m, mathematically this expression can be written as +300, the meaning of the number 300 from the “+” sign will not change. The distance or modulus of a number in mathematics is the same thing and can be written like this: |300|=300. The modulus sign of a number is indicated by two vertical lines.

And then we walked 200m in the opposite direction. Mathematically, we can write the return path as -200. But we don’t say “we walked minus two hundred meters,” although we returned, because distance as a quantity remains positive. For this purpose, the concept of a module was introduced in mathematics. You can write the distance or modulus of the number -200 like this: |-200|=200.
Module properties. Definition:

Modulus of a number or absolute value of a number

is the distance from the starting point to the destination point.

The modulus of an integer not equal to zero is always a positive number.
| The module is written like this:a

2. The modulus of a negative number is equal to the opposite number.
|- The module is written like this:a

3. The module of zero is equal to zero.
|0|=0

4. The modules of opposite numbers are equal.
| a|=|-The module is written like this:a

Questions on the topic:
What is the modulus of a number?
Answer: Modulus is the distance from the starting point to the destination point.

If you put a “+” sign in front of an integer, what happens?
Answer: the number will not change its meaning, for example, 4=+4.

If you put a “-” sign in front of an integer, what happens?
Answer: the number will change to, for example, 4 and -4.

Which numbers have the same modulus?
Answer: positive numbers and zero will have the same modulus. For example, 15=|15|.

Which numbers have the modulus of the opposite number?
Answer: for negative numbers, the modulus will be equal to the opposite number. For example, |-6|=6.

Example #1:
Find the modulus of numbers: a) 0 b) 5 c) -7?

Solution:
a) |0|=0
b) |5|=5
c)|-7|=7

Example #2:
Are there two different numbers whose moduli are equal?

Solution:
|10|=10
|-10|=10

The moduli of opposite numbers are equal.

Example #3:
Which two opposite numbers have modulus 9?

Solution:
|9|=9
|-9|=9

Answer: 9 and -9.

Example #4:
Follow these steps: a) |+5|+|-3| b) |-3|+|-8| c)|+4|-|+1|

Solution:
a) |+5|+|-3|=5+3=8
b) |-3|+|-8|=3+8=11
c)|+4|-|+1|=4-1=3

Example #5:
Find: a) the modulus of number 2 b) the modulus of number 6 c) the modulus of number 8 d) the modulus of number 1 e) the modulus of number 0.
Solution:

a) the modulus of the number 2 is denoted as |2| or |+2| It is the same.
|2|=2

b) the modulus of the number 6 is denoted as |6| or |+6| It is the same.
|6|=6

c) the modulus of the number 8 is denoted as |8| or |+8| It is the same.
|8|=8

d) the modulus of the number 1 is denoted as |1| or |+1| It is the same.
|1|=1

e) the modulus of the number 0 is denoted as |0|, |+0| or |-0| It is the same.
|0|=0