Expand Virazu module:

Section 2. Main characteristics.

Remaining solution – the union of the appearances of the okremikh parts:

x∈ (–∞; –3] ∪ [–1; + ∞).

Suggestion: x∈ (–∞; –3] ∪ [–1; + ∞) .

Rivnyannya mind | = | y |

butt 1(Algebra grade 8). Razv'yazati іz two modules: 2 * |x - 1| + 3 = 9 - | x - 1 |.

Solution:

Suggestion: x 1 = 3; x 2 = − 1.

butt 2(Algebra grade 8). Virishiti nerіvnіst:

Solution:

Rivnyannya mind | =y

butt 1(Algebra grade 10). Know x:

Solution:

It is also important to carry out a re-verification of the right part, otherwise you can write at the top of the pardon root. From the system it is clear that it is not possible to lie at the promenade.

Suggestion: x=0.

sumi module

Retail module

The absolute value of the difference between two numbers x that y dorivnyuє vіdstani between points with coordinates Xі Y on the coordinate line.

example 1.

butt 2.

Modulus of a negative number

To know the absolute value of a number, if it is less than zero, it is necessary to recognize it, as far as it is far from zero. Oscilki look forward to positive (it is impossible to go through negative steps, it’s just steps in a different direction), the result is always positive. Tobto,

To put it simply, the absolute value of a negative number may have an opposite value.

Zero module

Houses of power:

Why can't we say that an absolute value is a positive number: zero is neither negative nor positive.

Square module

The module of the square is always more advanced in terms of the square:

Attach graphics from the module

Often in tests and on tests, tasks are written down, yak it is possible to err, having less analyzed the graphs. Let's take a look at this task.

example 1.

Given a function f(x) = |x|. It is necessary to have a schedule for 3 to 3 times a week 1.

Solution:

Explanation: From the little one you can see that the graph is symmetrical along the Y axis.

butt 2. It is necessary to paint the graphs of the functions f(x) = |x–2| and g(x) = |x|–2.

Solution:

Explanation: the constant in the middle of the absolute value moves the entire graph right-handed, as it is negative, and left-hand, as positive. Ale postiyna zvnі peresuvatime graph uphill, which is more positive, i down, which is more negative (yak - 2 at the function g(x)).

Vertex coordinate x(Point, where two lines join, the top of the graph) - the whole number, on the chart, it moves to the left or to the right. A coordinate y- tse value, like the graph zsuvaetsya uphill chi down.

You can use such graphics for the help of online supplements for inspiration. With this help, you can at first wonder how to add constants to functions.

Interval method for tasks from module

The method of intervals is one of the best ways to know the difference in problems with a module, especially in the case of viruses.

To match the method, follow these steps:

1. Equal skin viraz to zero.
2. Know the meaning of the change.
3. Put points on the number line, subtract from point 2.
4. Signify on the spaces the sign of viraziv (negative or positive value) and paint the symbol - or + vіdpovіdno. The simplest is to assign a sign for the help of the substitution method (substituting whether the value is from the space).
5. Virishiti nerivnosti z otrimanimi signs.

butt 1. Virishiti path of intervals.

Solution:

One of the most interesting topics for uchnіv is rozvyazannya rivnyan, scho to avenge the change under the sign of the module. Let's take a look at the cob, why is it tied? Why, for example, are square equals more children clattering like peas, and with such far from the best understanding, like a module, can there be more problems?

In my opinion, all the folds are due to the existence of clearly formulated rules for the implementation of the module. So, virishyuyuchi square equal, learners know exactly what it is necessary for you to write down the formula of the discriminant, and then the formula of the root of the square equal. And what about the robit, how about the module being fixed on equal ground? We will try to clearly describe the necessary plan for the time, if it is equal to avenge the unknown under the sign of the module. We will bring a sprat of applications to the skin drop.

Ale for the cob module designation. Father, modulus of number a called the same number, like a I can't see -a, which is the number a less than zero. You can write it like this:

|a| = a if a ≥ 0 and |a| = -a, same as a< 0

Speaking about the geometrical sense of the module, the next thing to remember is that the skin decimal number has a single point on the numerical axis - її to coordinate. So the axis, the module or the absolute value of the number, is called the distance from the center of the point to the cob in the numerical axis. Always be given a positive number. In this order, the modulus of any negative number is positive. Before speech, learn at what stage a lot of learners start to stray. The module may have a number, and the result of the module's input is always a positive number.

Now let's move on without a hitch to the opening of the river.

1. Perspective equal to mind | \u003d s, de s - deisne number. The price can be checked for the help of the module.

All the current numbers are divided into three groups: t, which is greater than zero, t, which is less than zero, and the third group is the whole number 0. Let's write down the solution for the visual scheme:

(±c, if s > 0

Yakscho | x | = c, then x = (0, so c = 0

(no root, yakscho z< 0

1) | = 5, because 5> 0, then x = ±5;

2) | = -5 because -5< 0, то уравнение не имеет корней;

3) | = 0 then x = 0.

2. Equal to mind | f(x) | = b de b > 0. To complete this alignment, you need to get the module. Robimo tse: f(x) = b chi f(x) = -b. Now it is necessary to repair the skin from the otrimanih equals. Yakshcho at the weekend Rivnian b< 0, решений не будет.

1) | x + 2 | = 4, because 4 > 0, then

x + 2 = 4 or x + 2 = -4

2) | x 2 - 5 | = 11, because 11 > 0, then

x 2 - 5 = 11 or x 2 - 5 = -11

x 2 = 16 x 2 = -6

x = ± 4 no root

3) | x 2 - 5x | = -8, because -eight< 0, то уравнение не имеет корней.

3. Rivnyannya mind | f(x) | = g(x). For the replacement of the module, such a solution is equal to the mother, for example, part of the law is greater than zero, tobto. g(x) ≥ 0. Then we can calculate:

f(x) = g(x) or f(x) = -g(x).

1) | 2x - 1 | \u003d 5x - 10. The cost of matima is the root, yakscho 5x - 10 ≥ 0. The very beginning of the rozvyazannya of such rivnas.

1. O.D.Z. 5x – 10 ≥ 0

2. Solutions:

2x - 1 = 5x - 10 or 2x - 1 = - (5x - 10)

3. Combined O.D.Z. that decision, we take:

Root x = 11/7 is not suitable for O.D.Z., vin is less than 2, and x = 3 is your mind's satisfaction.

Suggestion: x = 3

2) | x - 1 | = 1 - x 2.

1. O.D.Z. 1 – x 2 ≥ 0

(1 – x)(1 + x) ≥ 0

2. Solutions:

x - 1 = 1 - x 2 or x - 1 = - (1 - x 2)

x 2 + x - 2 = 0 x 2 - x = 0

x = -2 or x = 1 x = 0 or x = 1

3. Joint decision and O.D.Z.:

More than a root is suitable x = 1 and x = 0.

Suggestion: x=0, x=1.

4. Equal to mind | f(x) | = | g(x)|. This is equal to two upcoming equals f(x) = g(x) or f(x) = -g(x).

1) | x 2 - 5x + 7 | = | 2x - 5 |. Tse equal equal to two will come:

x 2 - 5x + 7 = 2x - 5 or x 2 - 5x +7 = -2x + 5

x 2 - 7x + 12 = 0 x 2 - 3x + 2 = 0

x = 3 or x = 4 x = 2 or x = 1

Suggestion: x = 1, x = 2, x = 3, x = 4.

5. Rivnyannya, yakі vyrіshuyusya way of substitution (replace change). The Danish solution method is easiest to explain on a specific application. So, let's give a square equal to the module:

x 2 – 6|x| + 5 = 0. For the quality of the module x 2 = |x| 2 , that can be rewritten like this:

|x| 2 - 6 | x | + 5 = 0. Let's change | = t ≥ 0, then math:

t 2 - 6t + 5 \u003d 0. Considering the given equal, we assume that t \u003d 1 or t \u003d 5. Let's turn to replace:

|x| = 1 chi |x| = 5

x = ±1 x = ±5

Suggestion: x=-5, x=-1, x=1, x=5.

Let's look at another example:

x 2 + | – 2 = 0. For module quality x 2 = |x| 2 , to

|x| 2+ |x| - 2 = 0. Let's change | x | = t ≥ 0 then i:

t 2 + t - 2 \u003d 0. Virishingly given equal, it is acceptable, t \u003d -2 or t \u003d 1. Let's turn to replace:

|x| = -2 chi |x| = 1

No root x = ± 1

Suggestion: x=-1, x=1.

6. Another type of alignment is alignment with a "foldable" module. Up to such equalities one can see the equalization, in which there are modules in modules. Rivnyannya tsgogo mind can be violated, blocking the power of the module.

1) |3 – |x|| \u003d 4. D_yatimemo is the same, like in equals of another type. Because 4 > 0, then we take two equalities:

3 - | x | = 4 chi 3 – |x| = -4.

Now virazimo at the skin level module x, todi | = -1 chi |x| = 7.

Virishuemo skin from otrimanih equal. The first equal does not have a root, because -one< 0, а во втором x = ±7.

Verify x=-7, x=7.

2) | 3 + | x + 1 | | = 5

3 + | x + 1 | = 5 chi 3 + |x + 1| = -5

|x + 1| = 2 | x + 1 | = -8

x + 1 = 2 or x + 1 = -2. There is no root.

Suggestion: x=-3, x=1.

Іsnuє shchey and universal method of rozv'yazannya іvnyan іz module. Tse method of intervals. Ale mi yogo looked at.

blog.website, with a new or private copy of the material sent on the original binding.

The module is one of those silent speeches, about all the chules, but in truth, nobody normally understands. To this day there will be a great lesson, assignments to the top of the line from the modules.

I’ll tell you right away: the lesson will be awkward. І vzagalі modules vzagalі topic is notably clumsy. “So, obviously, awkward! My brain is growing! - to say a lot of learning, but all the brains are explored through those that most people have no knowledge at the head, but like crap. The first lesson is to turn crap into knowledge.

Trochy theory

So let's go. Let's start with the most important thing: what is a module? I'll guess that the modulus of the number is just the same number, but take it without the minus sign. Tobto, for example, $ \ left | -5 \right | = $5. Abo $\left | -129.5\right | = $129.5.

Is everything simple? Yes, simple. And why is the modulus of a positive number worth it? Here it is even simpler: the modulus of a positive number is equal to the number itself: $ \ left | 5\right | = $5; $\left| 129.5\right | = $129.5 etc.

Exit tsіkava rіch: different numbers can be the same module. For example: $ \ left | -5 \right|=\left| 5\right | = $5; $\left| -129.5 \right|=\left| 129.5\right | = $129.5. It doesn't matter if the numbers are the same for some modules: the numbers are the same. Also, it is important to note that the modulus of opposite numbers is equal:

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

Another important fact: the module is by no means negative. They took the number mi - even if it is positive, if it is negative - the yogo module will always be positive (or in the extreme it will be zero). For this reason, the modulus is often called the absolute value of a number.

In addition, since the module assignment for positive and negative numbers is combined, the module assignment for all numbers is taken globally. And itself: the modulus of the number is equal to the number itself, if the number is positive (or zero), or if the number is equal to the opposite number, if the number is negative. You can write down the same formula:

More is the modulus of zero, but vin zavzhdi is equal to zero. Moreover, zero is alone, as there is no opposite.

In this way, let's take a look at the $ y = \ left | x \right|$ and try to paint the її schedule, then we will see this “daw”:

Graph of the module and butt of perfection

From the bottom of the picture you can clearly see that $ \ left | -m \right|=\left| m \right|$, and the graph of the module does not go any lower than the x-axis. And yet, not everything: the straight line $y=a$ is marked with a red line, so with positive $a$ it gives us two roots: $((x)_(1))$ and $((x)_(2)) $, but let's talk about them later. :)

The cream of a purely algebraic design is more geometric. It is possible that there are two points on the number line: $((x)_(1))$ i $((x)_(2))$. І here viraz $\left| ((x)_(1))-((x)_(2)) \right|$ - don't just move between designated points. Abo, as always, a good vіdrіzka, scho zadnuє tsі points:

Of which designation is also pronounced, that the module is always negative. Let's move on to the right equals.

Basic formula

Well, harazd, they got out of the appointments. Ale didn’t feel any better. How to check the level, what should be done about this module itself?

Calm but calm. Let's start with the simplest speeches. Let's look at something like this:

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

So, add $x$ module 3. What can you add $x$ to? Well, judging by the appointment, we are totally in power $x=3$. Diyno:

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

What are the other numbers? Cap nibi pulls, scho є. For example, $ x = -3 $ - for the new $ \ left | -3 \right | = $3, then. necessary equanimity wins.

Then, perhaps, as a joke, think, do we know the numbers? And the axis is broken: there are no more numbers. Rivnyannia $ \ left | x \right|=3$ can only have two roots: $x=3$ and $x=-3$.

Now trochs can be put in order. Let the function $f\left(x \right)$ change under the modulo sign, and the right-handed substitute for the triplet can be set to a sufficient number $a$. We take equal:

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

Well, and how do you virishuvate? Guessing: $f\left(x \right)$ is quite a function, $a$ is a number. Tobto. vzagali be-yak! For example:

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

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

Best respect for each other is equal. One can say about the new eye: there is no root in the new one. Why? Everything is correct: what is needed in the new, so that the module is added to a negative number, which we never know, we already know that the module is a positive number, or zero in the extreme.

And the axis from the first equals is more fun. There are two options here: or under the sign of the module to be positive, and then $ \ left | 2x+1 \right|=2x+1$, otherwise ce viraz is still negative, and so $\left| 2x+1 \right|=-\left(2x+1 \right)=-2x-1$. At the first glance, our equal will be rewritten like this:

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

And it's easy to come out that the submodular virase $2x+1$ is effectively positive - to the number 5. That's it. we can calmly virishuvati tse rivnyannia - taking away the roots will be shmatkom vіdpovіdі:

Particularly distrustful, they can try to put the knowledge of the root of the vyhіdne equal and reconciliation, which will be a positive number for the module.

Now let's take a look at the negative submodular virusase:

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

Oops! I know everything clearly: we allowed that $2x+1 \lt 0$, and as a result took away that $2x+1=-5$ — ce viraz less than zero. Virishuemo otrimane equal, with whom you already know for sure that knowledge is the root of us:

At the same time, we took away the two returns again: $ x = 2 $ і $ x = 3 $. So, the total cost was three times greater, lower than the simple equal $\left| x \right|=3$, but nothing has changed. Then, perhaps, is there a universal algorithm?

So, such an algorithm is known. І at once mi yogo razberemo.

Zvіlnennya according to the sign of the module

Let's give us equal $ \ left | f\left(x \right) \right|=a$, and $a\ge 0$ (now, as we already know, there is no root). Then you can omit the modulus sign behind this rule:

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

In this rank, our alignment with the module splits into two, but even without a module. Axis and all the expansion! Let's try virishiti kіlka rivnyan. Let's get the axis out of this

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

Okremo razglyanaem, if the right is a dozen with a plus, and okremo if with a minus. Maemo:

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

From i all! They won two roots: $ x = $1.2 and $ x = -2.8 $. All solutions took literally two rows.

Ok, no food, let's take a look a little more seriously:

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

I am re-opening the module with plus and minus:

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

I'm starting a couple of rows - and the turnaround is ready! As I said, there is nothing collapsible in the modules. It's better to remember the sprat rules. To that we gave and proceeded to the right folding tasks.

Vipadok zminnoy right part

And now let's take a look at this equalization:

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

Tse equal in principle vіdrіznyaєtsya vіd pperednіh. Chim? And we, who are right-handed in the sign of equivalence of the cost of $2x$ - we cannot know for a long time which is more positive and which is negative.

How do you like this time? First, you need to understand once and for all what if the rights of a part of the equal to appear negative, then the equal is not a root- we already know that the modulus cannot equal a negative number.

And in a different way, if the right part is still positive (otherwise it is equal to zero), then you can work the same way as before: just expand the module with a plus sign and a minus sign.

In this way, we formulate a rule for additional 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.\]

We take away our jealousy:

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

Well, maybe $2x\ge 0$ we seem to be resting. To be honest, you can stupidly imagine the root, as we take away from the first equal, and reverse: what is the difference between chi and ni.

To that we will untie the very jealousy:

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

Well, yak z tsikh dvoh korenіv satisfying maybe $2x\ge 0$? So both! That's why you have two numbers: $ x = (4) / (3) \; $ i $ x = 0 $. Axis and all solutions.

I suspect that some of the students have already begun to feel bad? Well, let’s look at more folding:

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

Even if it looks malicious, in fact, all of them are equal to the “module of good functions” type:

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

And it turns out just like that:

\[\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.\]

From the nervousness of me, then we’ll figure it out - it’s like it’s supposed to be evil (it’s really simple, but we won’t be violating yoga). For the time being, let's take care of otrimanimy equals. We can see the first drop - if the module is opened with a plus sign:

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

Well, here I realized that it is necessary for all the brothers to be evil, to bring similar ones and marvel at what we see. And see the axis:

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

Blame the high multiplier $((x)^(2))$ for the shackle and take even simpler equal:

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

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

Here we were honored by the important power of creation, for the sake of which we laid out the rich term into multiples: twir is equal to zero, if you want one of the multipliers to be equal to zero.

Now we’ll figure it out for ourselves with other equals, what to enter when opening the module with a “minus” sign:

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

I know the same: tvir is equal to zero, if it is equal to zero, I want one of the multiples. Maemo:

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

Well, three roots were taken away from mi: $ x = 0 $, $ x = 1.5 $ i $ x = (2) / (3) \; Well, and what about the set of pide at vіdpovіd? For whom, let’s guess what we can doatkove obezhennya at the sight of unevenness:

How to vrahuvati tsiu vimogu? It is so easy to imagine that the root is found and it is reverifiable: there is a difference in the case of tsikh $x$ chi ni. Maemo:

\[\begin(align)& x=0\Rightarrow x-((x)^(3))=0-0=0\ge 0; \\& x=1,5\Rightarrow x-((x)^(3))=1,5-((1,5)^(3)) \lt 0; \\& x=\frac(2)(3)\Rightarrow x-((x)^(3))=\frac(2)(3)-\frac(8)(27)=\frac(10) (27)ge 0; \\end(align)\]

In this rank, the root $ x = $ 1.5 does not belong to us. I have only two roots:

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

Like a bachite, I didn’t have anything coherent in my mind - the equalization of the modules is always dependent on the algorithm. It is necessary to be better educated on the rich members and inconsistencies. So let's move on to folding tasks - there will already be not one, but two modules.

Alignment with two modules

Dosі mi vyvchali less than the simplest rіvnyannya - there's only one module and more. We corrected the "just now" into the other part of the unevenness, filed a module, so that the result was all equal to $ \ left | f\left(x \right) \right|=g\left(x \right)$ or go beyond the simple $\left| f\left(x \right) \right|=a$.

Ale, the childish garden is over - the hour has come to look more seriously. Let's look at this type:

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

The value of the mind "the module is the same as the module". A fundamentally important moment is the presence of other additions and multiples: only one left-handed module, another right-handed module - and nothing more.

Just think at once, that such equal variability is worse, lower those that we have learned and dosі. And the axis i nі: tsі rivnyannya virіshuyusya navіt simpler. Axis formula:

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

Mustache! We simply compare submodular virazi, putting a plus or minus sign in front of one of them. And then we will take away two equals - and the root is ready! Everyday dodatkovyh obmezhen, zhestnyh nerіvnosti just. Everything is simple.

Let's try this task:

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

Elementary, Watson! Opening modules:

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

Let's take a look at the skin vapadok:

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

The first equal has no root. Why if $3=-7$? For what values ​​of $x$? “What the fuck is $x$? Are you stoned? There’s not much $x$ in there,” you say. I will be right. We have obtained equivalence so that we cannot lay aside in the form of exchangeable $x$, and with this equivalence itself is wrong. That is why there is no root.

With other equals, all the trochs are cіkavіshe, but also more and more simply:

Like Bachimo, everything went literally in a couple of rows - we didn’t count the other line of line.

The result has a residual value: $ x = $1.

Well yak? Important? Obviously, no. Let's try again:

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

I know we are equal to the mind of $ \ left | f\left(x \right) \right|=\left| g\left(x \right) \right|$. To this, we immediately rewrite yoga, revealing the sign of the module:

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

It’s possible, someone asks at once: “Hey, what kind of lighthouse is this? Why is “plus-minus” on the right hand side, and not on the left side? Let me explain everything in a moment. In a good way, we are guilty of rewriting our equals like this:

Then we will need to open the arches, transfer all the additions into one line with the sign of parity (shards of parity, obviously, in both directions will be square), that and far away the root. But wait a minute: if “plus or minus” is standing in front of three dodanks (especially if one of them is a square viraz), you seem to look more folding, the situation is lower, if “plus or minus” is less likely to stand in front of two dodanks.

And yet, nothing matters to us to rewrite the day like this:

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

What happened? That is nothing special: they just remembered the lion and the right part of the missions. Dribnitsa, like a troch to forgive us life. :)

In the blink of an eye, it’s tse equal, looking at the options with plus and 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 equal root is $x=3$ and $x=1$. Another vzagalі є exact square:

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

There is only one root for this: $x=1$. Ale tse roots were already cut off earlier. In this order, the pіdsumkov vіdpovіd will have only two numbers:

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

Miss vikonan! You can take a pie from the police and get it. There are 2, your average.

Respectful respect. The presence of the same root with different variants of the module expansion means that the outer rich segments are divided into multiples, and the middle of these multiples will be bright. Diyno:

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

One of the module's powers: $ \left | acdot b \right|=\left| a \right|\cdot \left| b \right|$ (so that the module is more like the creation of modules), otherwise it can be rewritten like this:

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

Yak bachimo, we have the right vinic double multiplier. Now, in order to pick up all the modules from one side, you can blame the whole multiplier for the bow:

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

Well, now let's figure out what the addition is to zero, if we want one of the multipliers to reach zero:

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

In this rank, two modules were equal to two of the simplest equals, they talked about them at the beginning of the lesson. Such equalities are literally in a couple of rows.

Dane is respected, it is possible, to be superciliously foldable and unstoppable in practice. However, in truth, you can be informed where folded tasks are, lower those, as we can reasonably understand. Those modules can be combined with polynomials, arithmetic roots, and logarithms too. And in such situations, it is possible to reduce the burning riven by the path of guilt of something for the shackle, it can appear more and more than the river.

Now I would like to draw one more equal, as if at first glance I could look hazy. On the new "sticky" rich learners, navit te, yak vvazhayut, sho well sorted out in the modules.

Prote tse rіvnyannya vіrishuєtsya navіt simpler, lower those that we looked at earlier. As soon as you understand something, then you take one more trick to achieve a perfect match with the modules.

Otzhe, Rivnyanya:

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

Hі, tse not drukarska pardon: mizh modules itself is a plus. And we need to know, for such $x$ the sum of two modules is equal to zero. :)

Who has a problem? And the problem is that the skin module is a positive number, but in the extreme it is zero. How about adding two positive numbers together? Obviously, I am revisiting a positive number:

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

The rest of the row can be thought of: a single drop, if the sum of the modules is equal to zero, then the skin module is equal to 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 if the module is equal to zero? Only in one direction - if pіdmodulny vіraz dоrіvnyuє zero:

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

In this order, we have three points, in which the first module is reset to zero: 0, 1 and −1; and also two points, in which another module is set to zero: −2 and 1. However, it is necessary for us that both modules are set to zero at the same time, so among the known numbers it is necessary to choose t, which includes up to both sets. Obviously, there is more than one such number: $x=1$ — it will be a residual value.

splitting method

Well, we already looked at the couple of the day before and made up the impersonal receptions. Why do you think everything? And the axis i ni! At once we can look at the final reception - and at the same time the most important. Be aware of the splitting of the rivnyan іz module. What are you talking about? Let's turn back a little and look like a simple equal. For example, tse:

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

In principle, we already know how to act in such a way, because the standard construction of the form $\left| f\left(x \right) \right|=g\left(x \right)$. Ale try to marvel at the quality of the troch under another hood. More precisely, let's look at the viraz, what to stand under the sign of the module. I’ll guess that the modulus of any number can be equal to the number itself, or it can be the opposite of this number:

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

Vlasne, this ambiguity has the whole problem: the number of submodules changes (it’s worth it to change), it’s not clear to us whether it’s positive or negative.

But what if, on the other hand, vimagati, so that the number was positive? For example, let's say $3x-5 \gt 0$ - for which way we are guaranteed to take a positive number under the sign of the module, and which module itself can be called again:

In this rank, our zeal to pretend to be on a linear line, as it is easy to swear:

It's true, all you think about it is more than enough to think $3x-5\gt 0$ - we ourselves managed to help you, so that the module can be unambiguously unlocked. So let's put the knowledge of $x=\frac(5)(3)$ into the mind and reverb it:

To go out, our help does not win over the assigned value of $x$, because Viraz appeared to be equal to zero, but we need it to be strictly greater than zero. Zhurbinka. :(

Ale no big deal! Another option is $3x-5 \lt 0$. More than that: another point $3x-5=0$ — it is necessary to look at it this way, otherwise the decision will be incomprehensible. Let's take a look at the $3x-5 \lt 0$ vipadok:

Obviously, the module is marked with a minus sign. But then again, the situation is wondrous: I’m left-handed, and right-handed at the same viraz:

Tsikavo, with such $x$, $5-3x$ will be more expensive than $5-3x$? In the presence of such equals, the Captain is obvious, choking on his heel, but we know: the ceremony is equal to that, tobto. vono vіrne for whatever the meaning of the change!

And tse means that we are ruled by $x$. Vodnocha we have є obmezhennya:

In other words, it will not be a short number, but a whole interval:

Nareshti lost one more perspective: $3x-5=0$. Everything is simple here: the modulus will be zero, and the modulus of zero will be equal to zero (it’s not directly pronounced):

Ale todі vhіdne rіvnyannya $ \ left | 3x-5 \right|=5-3x$ rewrite like this:

This root was already taken higher, if we looked at the $3x-5\gt 0$ drop. More than that, the price of the root and the solutions equal $3x-5=0$ - the value of the exchange, as we ourselves entered, to reset the module.

In this order, the crim interval is the dominant number for us, which lies on the very end of the interval:

Residual proof: $x\in \left(-\infty ;\frac(5)(3) \right]$ It's not too loud to make such crap in the widget until it's simple (essentially - linear) alignment with the module Well, please : this is why the folding of the module is due to the fact that in such equalities it can appear absolutely unreproducible.

Where else is more important: we have carefully developed a universal algorithm for solving the problem with the module! І the whole algorithm is formed from the next steps:

1. Equate the skin module, which is equal, to zero. We take away sprats of equal;
2. Put all the numbers equal and put the root on the number line. As a result, a direct rise to a sprinkling of intervals, on the skin of any of the modules, unambiguously rises;
3. Virishiti vihіdne іvnyannja for dermal interval and ob'єdnati otrimaniі vіdpovіdі.

From i all! Less than one food is left: where do you go your own roots, cut off on the 1st crochet? Suppose we have two roots: $ x = 1 $ i $ x = 5 $. The stench rose numerically straight for 3 pieces:

Well, what are the intervals here? I realized that there are three of them:

1. Naylivishy: $x \lt 1$ — the single element itself is not included in the interval;
2. Central: $1\le x \lt 5$ - the axis here is one per interval to enter, prote not to enter five;
3. The right one: $x\ge 5$ - Five days to get in here!

I guess you already understood the law. The leather interval includes the left end and does not include the right.

At first glance, such a record may appear unhandled, illogical, and seem to be hazy. Ale turn: after a little training, you will discover that such a pidkhid itself is the most superior and, for that matter, does not unequivocally develop modules. It’s better to win such a scheme, then think twice: take a left / right turn at the current interval, or throw yoga on the offensive.

On which lesson will end. Take charge of the task for self-sufficiency, train, compete with influences - and we will work in the coming lesson, which will be assigned to the nervousness of the modules.