Operations with roots. Apply as a virishuvati to the roots. Now I'll be on my own

⁰

$\sqrt(a)=b$, so $b^2=a$, de $a≥0,b≥0$

Apply:

$\sqrt(49)=7$, because $7^2=49$
$\sqrt(0,04)=0,2$,scales $0,2^2=0,04$

How to take the square root of a number?

In order to find the square root of the number, you need to put your own power: how should the number be given to the square by the root?

For example. Use the root: a) (sqrt (2500)); b) $\sqrt(\frac(4)(9))$; c) $\sqrt(0.001)$; d) $\sqrt(1\frac(13)(36))$

a) What is the number of the square to give (2500)?

$\sqrt(2500)=50$

b) What is the number of the square to give $\frac(4)(9)$?

$\sqrt(\frac(4)(9))$ $=$$\frac(2)(3)$

c) What is the number of the square, give (0.0001)?

$\sqrt(0.0001)=0.01$

d) What is the number squared to give $\sqrt(1\frac(13)(36))$? In order to give an opinion on the supply, it is necessary to translate it to the wrong one.

$\sqrt(1\frac(13)(36))=\sqrt(\frac(49)(16))=\frac(7)(6)$

Respect: Want $-50$, $-\frac(2)(3)$ , $-0,01$,$- \frac(7)(6)$ can also match power supply , but do not protect them, because the square root is always positive.

Head power of the root

As you can see, in mathematics, be-such a diї є zvorotne. At the addition - vіdnimannya, at the plurality - rozpodіl. Zvorotne diyu zvedennya square - forcing the square root. That tі dії compensate one for one:

$(\sqrt(a))^2=a$

Tse and є the head power of the root, as it is most often victorious (including those in the ODE)

Butt . (Head of the ODE). Find the value of the virus $\frac((2\sqrt(6))^2)(36)$

Solution :$\frac((2\sqrt(6))^2)(36)=\frac(4 \cdot (\sqrt(6))^2)(36)=\frac(4 \cdot 6)(36 )=\frac(4)(6)=\frac(2)(3)$

Butt . (Head of the ODE). Find the value of the virus $(\sqrt(85)-1)^2$

Solution:

Suggestion: $86-2\sqrt(85)$

Obviously, when working with a square root, it is necessary to vicorate more.

Butt . (Head of the ODE). Get Virase Value $5\sqrt(11) \cdot 2\sqrt(2)\cdot \sqrt(22)$
Solution:

Suggestion: $220$

4 rules about how to forget forever

The root does not start to grow

Butt: $\sqrt(2)$,$\sqrt(53)$,$\sqrt(200)$,$\sqrt(0,1)$ etc. - Vityagti korіn z number not zavzhd it is possible and it's okay!

The root of the number, the same number

It is not required to enter $\sqrt(2)$, $\sqrt(53)$, especially. Tse numbers, that are not numbers, so, but not everyone in our world lives in numbers.

The root is hovering over unknown numbers

Therefore, in your assistants, do not add such records $\sqrt(-23)$,$\sqrt(-1)$, etc.

The time has come for the rose ways of rooting. The stench is based on the power of roots, zokrema, on equivalence, as it is fair for any number b.

Below, we will look at the main ways of rooting.

Let's start from the simplest point - from the study of the roots from the natural numbers from the variations of the tables of squares, the tables of cubes too.

Like tables of squares, cubes too thin. if you don’t know how to use your hands, then it’s logical to use the method of foraging the root, which you can use to lay out the sub-root number on simple multipliers.

Okremo varto zupinitsya on those that are possible for the root of unpaired displays.

Nareshti, let's look at a way that allows you to successively shukat the value of the root.

Let's get started.

Vicoristanna tables of squares, tables of cubes too.

In the simplest ways, roots can be obtained by tables of squares, cubes only. What are the tables?

Table of squares of integers from 0 to 99 inclusive (shown below) and two zones. The first zone of the table is arranged on a gray background, for the additional choice of a song row and a song column, it allows you to add up the number of entries from 0 to 99. For example, we chose a row of 8 tens and a row of 3 singles, and fixed the number 83. Another zone borrows part of the table that is left out. The skin її komіrka is located on the pereline of the sing row and the sing stitch, and avenge the square of the general number from 0 to 99. On the edge of the row we have chosen, 8 tens and 3 singles, there is a middle with the number 6889, as well as the square of the number 83.

Tables of cubes, tables of fourth steps of numbers from 0 to 99 are similar to tables of squares, only in another zone there are cubes, quarter steps are thin. random numbers.

Tables of squares, cubes, quarter steps, too. allow you to draw square roots, cubic roots, roots of the fourth degree, etc. it is clear from the numbers that they have in their tables. Let's explain the principle of their stosuvannya pіd hour of rooting.

Suppose we need to find the root of the n-th level of the number a, and the number a is placed in the tables of the n-th levels. According to this table, the number b is known so that a = b n. Todi , Otzhe, the number b will be the root of the n-th step.

Like an example, it’s shown, like a cubic root from 19683. We know the number 19683 in the table of cubes, we know that this number is the cube of the number 27, .

It was clear that the tables of n-their steps are more efficient when the roots are forgotten. Prote їх often do not show up under the hands, like folding for the singing hour. Moreover, it is often necessary to take the root of the numbers, so as not to miss the same tables. In these experiences, one should be led to other methods of extracting the root.

Decomposition of the root number into simple multipliers

Dose in a manual way, which allows you to carry out the extraction of the root from a natural number (as it is, the root of the natural number grows out) - the distribution of the root number into simple multipliers. Yogo the essence of the field is in the offensive: if you do it easy to pay taxes at the sight of the step with the necessary show, which allows you to take the value of the root. Let's explain the moment.

Let the root of the n-th degree be drawn from the natural number a, the i-th value is one b. And here the equality a=b n is correct. The number b, as if it were a natural number, can be represented by looking at the addition of all its prime factors p 1 , p 2 , …, pm by looking at p 1 2 · ... · pm) n . Scattering the numbers on the simple multipliers is the same, then the spreading of the root number a on the simple multipliers looks like (p 1 ·p 2 ·...·p m) n , which allows you to calculate the value of the root yak.

Respectfully, that the th layout on the prime factors of the root number a can be represented as (p 1 ·p 2 ·…·p m) n, then the root of the n-th step of the th number a does not rise at all.

Let's take a look at the time of the ceremony of application.

butt.

Take the square root of 144.

Solution.

If you turn back to the table of squares given in the front paragraph, then it is clearly visible that 144 \u003d 122, the stars realized that the square root of 144 is 12.

Ale, at the light of this point, cackle us, as the root for the help of the laying of the root number 144 in simple multipliers. Let's figure out the way of rozvyazannya.

Rozklademo 144 simple multipliers:

Tobto, 144 = 2 2 2 2 3 3 . On the basis of the removed layout, you can carry out the following transformation: 144 = 2 2 2 2 3 3 = (2 2) 2 3 2 = (2 2 3) 2 = 12 2. Otzhe, .

Vykoristovuyuchi power level and root, solution can be issued and trohi іnakshe:.

Suggestion:

For fixing the material, we look at the solution of two more applications.

butt.

Calculate the value of the root.

Solution.

The layout of the simple multipliers of the root number 243 may look like 243 = 35. In this order, .

Suggestion:

butt.

Chi is the value of the root as a whole number?

Solution.

Schob vodpovisti on the chain of food, spread out the root number on simple multipliers and marvel at what can be imagined in the cube of an integer number.

Maemo 285 768 = 2 3 3 6 7 2 . Otrimane layout does not seem like a cube of an integer, the steps of a simple multiplier 7 are not a multiple of three. Also, the cube root from the number 285768 does not rise on a national level.

Suggestion:

Ni.

Explanation of roots from fractional numbers

The hour has come to rise, as the roots emerge from the fractional number. Let the fractional root number be written as p/q. Vidpovidno to the power of the root is often just offensive jealousy. Z ієї rivnostі vyplivaє root rule for a fraction: the root from the shot is the most precious part of the root from the number book to the root from the banner.

Let's take a look at the butt of the forging of the root from the shot.

butt.

Why is the square root of the great fraction 25/169 worth?

Solution.

According to the table of squares, it is known that the square root from the numeral book of the last fraction is 5, and the square root from the banner is 13. Todi . On which root of the great fraction 25/169 is completed.

Suggestion:

The root of the tenth fraction of a mixed number is drawn up after replacing the root numbers with strong fractions.

butt.

Let's take the cube root from the tenth fraction 474.552.

Solution.

Let's give a decimal decimal fraction to a supernumerary fraction: 474.552 = 474552/1000. Todi . It has lost its vitality of cubic roots, which are found in the number book and banner of the cut shot. so yak 474 552=2 2 2 3 3 3 13 13 13=(2 3 13) 3 =78 3 i 1 000=10 3 then і . Lost to complete the calculation .

Suggestion:

Reversing the root of a negative number

Okremo varto zupinitsya on the extraction of the root of negative numbers. When rooted, we said that if the sign of the root is an unpaired number, then under the sign of the root it can be a negative number. Such records were given an offensive sense: for a negative number −a and an unpaired exponent of the root 2 n−1, we have . Tsya jealousy give the rule of finding the roots of the unpaired degree of negative numbers: in order to extract the root from a negative number, it is necessary to extract the root from the opposite number of a positive number, and put a minus sign before taking the result.

Let's take a look at the solution.

butt.

Find the meaning of the root.

Solution.

Let's remake the vihіdny viraz, so that under the sign of the root a positive number appeared: . Now the number is replaced by the greatest fraction: . Zastosovuєmo rule for the extraction of the root from the great fraction: . It was lost to count the root in the number book and the banner of the taken fraction: .

Let's make a short note of the solution: .

Suggestion:

Bitwise value of the value of the root

The number is changed at the root under the root, as if with the help of the other one, the other people are taken in at the sight of the n-th degree of this number. Ale, with whom, it is necessary to know the meaning of the root, wanting to be accurate to the singing sign. In this way, for removing the root, you can speed up the algorithm, which allows you to sequentially take a sufficient number of values of the rows of the shuk number.

At the first stage of this algorithm, it is necessary to determine what the most significant digit of the root value is. For which one, it is necessary to sequentially add to the steps n of the number 0, 10, 100, ... until the moment when the number is taken away, which shifts the sub-root number. The same number, like we were called to step n on the front stage, is the same senior rank.

For example, let's look at the algorithm for the first time of learning the square root of five. We take the numbers 0, 10, 100, ... and square them, subtract the number that we move 5. May 02 = 0<5 , 10 2 =100>5, later, the senior rank will be the rank of singles. The value of this order, as well as the younger ones, will be found on the upcoming steps of the root forcing algorithm.

All advances to the algorithm may be based on the last clarified value of the root for the calculation of what the values of the upcoming rows of the shukan value of the root are known to be, starting from the eldest and extending to the youngest. For example, the value of the root for the first crochet is 2, the other - 2.2, the third - 2.23, then 2.236067977 .... We will describe how the significance of the discharges is shown.

Znakhodzhennya discharges are carried out with an additional enumeration of all possible values 0, 1, 2, ..., 9 . For whom, in parallel, the n-th steps of the second numbers are counted, and the stinks are ranked with the root number. If at some stage the value of the step is to reverse the number of the root, then the value of the order, which is similar to the forward value, is important to know, and a transition is made to the next step of the algorithm of the root, which is not considered, then the value of this order is more expensive.

It is understandable qi moments on the same example of the study of the square root of five.

On the back, we know the value of the singles. Iterate over the values 0, 1, 2, …, 9 , enumerate the values 0 2 , 1 2 , …, 9 2 doti, dots obsessively, greater than the root number 5 . All calculations must be submitted manually in the following tables:

So the value of a row of singles is more expensive 2 (oskіlki 2 2<5 , а 2 3 >five). Let's move on to the significance of the order of ten. With which we will square the numbers 2.0, 2.1, 2.2, ..., 2.9, taking into account the subtraction of the value with the root number 5:

Oskilki 2.2 2<5 , а 2,3 2 >5, then the value of the order of tenths is 2. You can go to the value of the order of hundreds:

So the same value of the root of five was found, it is equal to 2.23. And so you can continue to know the meaning: 2,236, 2,2360, 2,23606, 2,236067, … .

For fixing the material, we will analyze the root extract to the nearest hundred for the help of the examined algorithm.

I’m going to sign the senior rank. For whom, the numbers 0, 10, 100 and so on are reduced to a cube. docks take the number to change 2 151.186 . May 03 =0<2 151,186 , 10 3 =1 000<2151,186 , 100 3 =1 000 000>2 151.186 in this rank, the senior rank is the rank of tens.

Significantly yoga meaning.

Oskilki 10 3<2 151,186 , а 20 3 >2 151.186 , then the value is in the order of dozens more 1 . Let's go to one.

Otzhe, the value of the row of singles is more expensive 2. Let's go to ten.

Oskіlki navit 12.9 3 less for the root number 2 151.186, then the value in the order of ten is more expensive 9. The rest of the algorithm has lost its viconate, it will give the value of the root with the necessary accuracy.

At this stage, the root value was found to the nearest hundredth: .

At the end of this article, I would like to say that there are no other ways to improve the roots. Ale, for the greater part, the zavdan is quite quiet, as if we had grown bigger.

List of literature.

Makarichev Yu.M., Mindyuk N.G., Neshkov K.I., Suvorova S.B. Algebra: tutor for 8 cells. lighting installations.
Kolmogorov A.M., Abramov A.M., Dudnitsin Yu.P. that іn Algebra and cob analysis: Handyman for 10 - 11 classes of lighting installations.
Gusev V.A., Mordkovich A.G. Mathematics (a guide for students to technical colleges).

Wait, cats! Last time, we reportedly figured out what the root is (if you don’t remember, I recommend reading it). The head visnovok of that lesson: there is only one universally designated root, as you need to know. Reshta - nonsense that marnuvannya hour.

We have gone today. Vchimosya multiply the root, vivimo deaki problems, povyazanі z multiplications (as if the problem is not virishit, then in sleep the stench can become fatal) and as a consequence we will work out. So stock up on popcorn, be smarter - and we'll fix it.

Haven't you smoked Aja Vi yet?

To complete the lesson of viishov great, I divided yoga into two parts:

On the back of our hand, we will analyze the rules of multiplication. Kep yak natyakaє: if there are two roots, between them there is a sign “multiply” - and we want to work out of them.
Then let's analyze the situation: there is one great root, and it was hard for us to show it at the sight of two simpler roots. For some kind of perelaku tse buvaє is necessary - except for food. Let's take a look at the algorithm.

Tim, who can't wait to get to the next part, you're welcome. Let's fix it in order.

Basic multiplication rule

Let's start with the simplest - the classic square root. Tі samі, yaki are denoted by $\sqrt(a)$ and $\sqrt(b)$. For them, everything is obvious:

Multiplication rule. To multiply one square root by another, you just need to multiply their sub-roots of the virazi, and write the result under the radical radical:

\[\sqrt(a)\cdot \sqrt(b)=\sqrt(a\cdot b)\]

The annual supplementary borders on the numbers that stand right-handed or left-handed do not overlap: as root multipliers are used, then th tvir tezh іsnuє.

apply. Let’s take a look at the chotiri applied with numbers:

\[\begin(align) & \sqrt(25)\cdot \sqrt(4)=\sqrt(25\cdot 4)=\sqrt(100)=10; \\ & \sqrt(32)\cdot \sqrt(2)=\sqrt(32\cdot 2)=\sqrt(64)=8; \\ & \sqrt(54)\cdot \sqrt(6)=\sqrt(54\cdot 6)=\sqrt(324)=18; \\ & \sqrt(\frac(3)(17))\cdot \sqrt(\frac(17)(27))=\sqrt(\frac(3)(17)\cdot \frac(17)(27 ) ))=\sqrt(\frac(1)(9))=\frac(1)(3). \\ \end(align)\]

Like Bachite, the main sense of the rule is the forgiveness of irrational viruses. І as in the first butt, mi b i sam_ vityagli root z 25 i 4 without new new rules, then let the coat begin: $ \ sqrt (32) $ і $ \ sqrt (2) їх tvіr appears as an exact square, to that the root of the new rational number.

Okremo want to bi signify the remaining row. There, insults are rooted, virazi and fractions. Zavdyaks create a lot of multipliers quickly, and the whole viraz transforms into an adequate number.

Zvichayno, do not start everything will be so garno. Sometimes, under the roots, we stand at the same time crap - it’s unreasonable that you work with her and how to remake the following multiplication. Trohi later, if you remember the irrational equality and nervousness, there will be signs of change in that function. And even more often, stacking the manager is a lakraz and repaying for those who show up like warehouses, who are quick, or multipliers, after which the task of bagator is to say goodbye.

In addition, it is necessary to multiply the two roots in a language that is not necessary. You can multiply once by three, chotiri - that even ten! The rule of the kind does not change. Take a look:

\[\begin(align) & \sqrt(2)\cdot \sqrt(3)\cdot \sqrt(6)=\sqrt(2\cdot 3\cdot 6)=\sqrt(36)=6; \\ & \sqrt(5)\cdot \sqrt(2)\cdot \sqrt(0.001)=\sqrt(5\cdot 2\cdot 0.001)= \\ & =\sqrt(10\cdot \frac(1) (1000))=\sqrt(\frac(1)(100))=\frac(1)(10). \\ \end(align)\]

I again have little respect for another butt. Like a bachite, the third multiplier has a tenth root at the root - in the process, we calculate it by replacing it with the most significant one, since everything is easy to shorten. So the axis: I also recommend avoiding tens of fractions in any irrational virazes (to take away one radical badge). At the future, to spare you a couple of hours and nerves.

Ale tse buv lyrical vіdstup. Now let's look at the deepest twist - if the number of $n$ stands in the show of the root, and not just the "classic" two.

Vipadok of a happy show

Otzhe, with square roots, they got it. And what about working with cubic ones? What did you get at the roots of a good step $n$? But all the same. The rule is overridden by ourselves:

To multiply two roots of the $n$ step, it is sufficient to multiply the sub-roots of the virazi, after which the result can be written under one radical.

Zagalom nothing complicated. Hiba scho could be calculated more. Let's take a sprat of applications:

apply. Calculate create:

\[\begin(align) & \sqrt(20)\cdot \sqrt(\frac(125)(4))=\sqrt(20\cdot \frac(125)(4))=\sqrt(625)= five; \\ & \sqrt(\frac(16)(625))\cdot \sqrt(0,16)=\sqrt(\frac(16)(625)\cdot \frac(16)(100))=\sqrt (\frac(64)(((25)^(2))\cdot 25))= \\ & =\sqrt(\frac(((4)^(3)))(((25)^(3 ) ))))=\sqrt(((\left(\frac(4)(25) \right))^(3)))=\frac(4)(25). \\ \end(align)\]

I renew the respect of my friend viraz. We multiply the cubic root, let go of the decimal fraction, and as a result, we take additional numbers 625 and 25 into the banner.

To that we simply saw the exact cube at the numeral and banner, and then we scurried along one of the key authorities (or, as always, appointed) of the root of the $n$-th step:

\[\begin(align) & \sqrt(((a)^(2n+1)))=a; \\ & \sqrt(((a)^(2n)))=\left| a\right|. \\ \end(align)\]

Similar "scams" can spare you an hour to sleep or control robots, remember that:

Do not hurry to multiply the numbers in the root expression. Back to back: raptom is there "encrypted" the exact steps of what a viraz?

For all the obviousness of this respect, I can be recognized, most of the unprepared scientists did not succumb to the exact steps. The stench of the stench multiplies everything ahead, and then we wonder: why did you have such crazy numbers? :)

Vtіm, all the children's belkіt at the porіvnіnі z tim, scho mi vivchimo infection.

Reproduction of roots with various indications

Well, good, now we can multiply the root with the same show-offs. And what are the show-offers of raznі? Let's say, how do you multiply the ultimate $\sqrt(2)$ by some crap like $\sqrt(23)$? Can you work hard?

So obviously you can. Everyone fights the axis for the qiєyu formula:

Root multiplication rule. To multiply $\sqrt[n](a)$ by $\sqrt[p](b)$, it's enough for the vikonati to transform:

\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n)))\]

However, the formula is more practical than mind you subroots virazi nevid'emni. Tse more important respect, as we turn a little bit later.

In the meantime, let's take a look at a couple of applications:

\[\begin(align) & \sqrt(3)\cdot \sqrt(2)=\sqrt(((3)^(4))\cdot ((2)^(3)))=\sqrt(81 8) = sqrt(648); \\ & \sqrt(2)\cdot \sqrt(7)=\sqrt(((2)^(5))\cdot ((7)^(2)))=\sqrt(32\cdot 49)= \sqrt(1568); \\ & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(625\cdot 9)= \sqrt(5625). \\ \end(align)\]

Like bachite, nothing folding. Now let's sort it out, the stars have taken the help of invisibility, and what will be, as if we are destructible.

It is not easy to multiply the root

Why do the roots of the virazi seem to be invisible?

Obviously, you can become like a school reader and with a reasonable look, quote a friend:

In addition, the non-negativity of the relationship with different designations of the roots of the paired and unpaired steps (apparently, the areas of designation of the stinks are also different).

Well, what became clearer? I especially, if I read the 8th class censure, I understood for myself approximately the following: not sensible. :)

So I will explain everything in a normal way.

On the back, for sure, the stars in a flash, the multiplication formula is taken, it is pointed higher. For whom I will guess one important power of the root:

\[\sqrt[n](a)=\sqrt(((a)^(k)))\]

In other words, we can calmly raise the root of the root at any natural step $k$ - with which the indicator of the root happens to be multiplied by the same step. Otzhe, we can easily be known as a root to a sacrilegious display, after which we multiply. Zvіdsi i take the formula of the plural:

\[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p)))\cdot \sqrt(((b)^(n)))= \sqrt(((a)^(p))\cdot ((b)^(n)))\]

But there is one problem, which is sharply between the blockage of these formulas. Let's look at this number:

Apparently, we can add to the induced formula, whether it be the world. Let's try to add $k=2$:

\[\sqrt(-5)=\sqrt(((\left(-5 \right))^(2)))=\sqrt(((5)^(2)))\]

The minus was taken away by the very one that the square burns the minus (as if it were another boy's step). And now we’re going to turn around: we’ll speed up a double in a show and a step. Adzhe be-yak equanimity can be read as levoruch-right, so right-handed-levoruch:

\[\begin(align) & \sqrt[n](a)=\sqrt(((a)^(k)))\Rightarrow \sqrt(((a)^(k)))=\sqrt[n ](a); \\ & \sqrt(((a)^(k)))=\sqrt[n](a)\Rightarrow \sqrt(((5)^(2)))=\sqrt(((5)^( 2))) = sqrt(5). \\ \end(align)\]

And then go out like crap:

\[\sqrt(-5)=\sqrt(5)\]

What can't be, that's $\sqrt(-5) \lt 0$, but $\sqrt(5) \gt 0$. Also, for the pair steps and negative numbers, our formula does not work. Why do we have two options:

Get on the wall to state that mathematics is a stupid science, that there are rules, but it’s still inaccurate;
Introduce additional exchanges, for which the formula will become 100% working.

In the first variant, we happen to constantly wag "unpractical" depressions - it's important, for a long time and fu. That is why mathematicians gave priority to another option. :)

Ale don't worry! In practice, the exchange does not in any way contribute to the calculation, to the fact that all these problems are more than the roots of an unpaired level, and minuses can be blamed for them.

Therefore, we formulate one more rule, as if to expand in front of all the roots:

First, multiply the roots, grow in such a way that the root roots of the viraz are negative.

butt. In the middle of $\sqrt(-5)$ you can use minus z-n_d of the root sign - then everything will be fine:

\[\begin(align) & \sqrt(-5)=-\sqrt(5) \lt 0\Rightarrow \\ & \sqrt(-5)=-\sqrt(((5)^(2))) =-\sqrt(25)=-\sqrt(((5)^(2)))=-\sqrt(5) \lt 0 \\ \end(align)\]

Do you know the difference? If you lose the minus under the root, then when you square the root virase into the square of the root, you will get crap. And if you blame a minus on the back, then you can want to make / clean up the square until the blue - the number will become negative.

In this order, the most correct and the least possible way to multiply the roots of attacks:

Take away all the minuses of the radical radicals. The minuses are only in root unpaired multiplicity - they can be put before the root i if speed is necessary (for example, there are two of these minuses).
Vikonati multiplied sgіdno s rules, razglyanym more at this day's lesson. As the indicators of the root are the same, we simply multiply the root of the virazi. And even more different - victoriously evil formula \[\sqrt[n](a)\cdot \sqrt[p](b)=\sqrt(((a)^(p))\cdot ((b)^(n) ))\].
3. Enjoy the result and good marks. :)

Well, what? Shall we work out?

Example 1. Forgive Viraz:

\[\begin(align) & \sqrt(48)\cdot \sqrt(-\frac(4)(3))=\sqrt(48)\cdot \left(-\sqrt(\frac(4)(3 ) )) \right)=-\sqrt(48)\cdot \sqrt(\frac(4)(3))= \\ & =-\sqrt(48\cdot \frac(4)(3))=- \sqrt(64)=-4; \end(align)\]

The simplest option: the indicators of the root are the same and unpaired, the problem is less in the minus of another multiplier. We blame this minus nafig, if it is easy to get into it.

Example 2. Forgive Viraz:

\[\begin(align) & \sqrt(32)\cdot \sqrt(4)=\sqrt(((2)^(5)))\cdot \sqrt(((2)^(2)))= \sqrt(((\left(((2)^(5)) \right))^(3))\cdot ((\left(((2)^(2)) \right))^(4) ))= \\ & =\sqrt(((2)^(15))\cdot ((2)^(8)))=\sqrt(((2)^(23))) \\ \end( align)\]

Here those who came out with an irrational number would be rich. So, so buvaє: we couldn’t get the root right, but we took the suttavo and asked the viraz.

Example 3. Forgive Viraz:

\[\begin(align) & \sqrt(a)\cdot \sqrt(((a)^(4)))=\sqrt(((a)^(3))\cdot ((\left((( (a)^(4)) \right))^(6)))=\sqrt(((a)^(3))\cdot ((a)^(24)))= \\ & =\sqrt ( ((a)^(27)))=\sqrt(((a)^(3\cdot 9)))=\sqrt(((a)^(3))) \end(align)\]

Axis tse zavdannya hotіv b revnut your respect. There are two points here:

Under the roots, there is not a specific number but steps, but a change of $a$. At first glance, it’s not clear at first glance, but in reality, when solving mathematical tasks, it’s most likely that the mother will change on the right herself.
For example, we contrived to speed up the demonstration of the root and steps of the root of the expression. Such traplyaetsya dosit often. І tse means that it is possible to prostrate the calculation, otherwise it will be the main formula.

For example, you could do it like this:

\[\begin(align) & \sqrt(a)\cdot \sqrt(((a)^(4)))=\sqrt(a)\cdot \sqrt(((\left(((a)^( 4)) \right))^(2)))=\sqrt(a)\cdot \sqrt(((a)^(8))) \\ & =\sqrt(a\cdot ((a)^( 8)))=\sqrt(((a)^(9)))=\sqrt(((a)^(3\cdot 3)))=\sqrt(((a)^(3))) \ \ \end(align)\]

As a matter of fact, these transformations were more than just another radical. If you do not write down in detail all the interim lines, then as a result, the total will be significantly reduced.

In fact, we were already sticking with similar tasks, if they were violating the butt of $\sqrt(5)\cdot \sqrt(3)$. Now yoga can be painted in a much simpler way:

\[\begin(align) & \sqrt(5)\cdot \sqrt(3)=\sqrt(((5)^(4))\cdot ((3)^(2)))=\sqrt(( (\left(((5)^(2))\cdot 3 \right))^(2)))= \\ & =\sqrt(((\left(75 \right))^(2))) =Sqrt(75). \end(align)\]

Well, well, with multiple roots, they rose. Now let's take a look at the reversal operation: what work, if you stand under the roots?

fact 1.
$\bullet$ Assuming I don't know the number $a$ (that's $a\geqslant 0$). Todi (arithmetic) square root from the number \ (a \) is called such an unknown number \ (b \), when squaring it, we take the number \ (a \): \[\sqrt a=b\quad \text(the same as )\quad a=b^2\] Wow $a\geqslant 0, b\geqslant 0$. Qi obmezhennya є important mindful base of the square root and the next memory!
Guess what, be it a number squared, gives an indeterminate result. So $100^2=10000\geqslant 0$ and $(-100)^2=10000\geqslant 0$.
$\bullet$ Why $\sqrt(25)$? We know that $5^2=25$ and $(-5)^2=25$ . Since we can know a non-negative number, then $-5$ does not fit, then, $\sqrt(25)=5$ (sparks $25=5^2$ ).
The value of $\sqrt a$ is called the variation of the square root of the number $a$ , and the number $a$ is called the sub-root of the number $a$ .
$\bullet$ Vihodyachi z vzachennya, virazu $\sqrt(-25)$ , $\sqrt(-4)$ і etc. no sense.

fact 2.
For quick calculations, it will be appropriate to calculate the table of squares of natural numbers from $1$ to $20$: \[\begin(array)(|ll|) \hline 1^2=1 & \quad11^2=121 \\ 2^2=4 & \quad12^2=144\\ 3^2=9 & \quad13 ^2=169\\ 4^2=16 & \quad14^2=196\\ 5^2=25 & \quad15^2=225\\ 6^2=36 & \quad16^2=256\\ 7^ 2=49 & \quad17^2=289\\ 8^2=64 & \quad18^2=324\\ 9^2=81 & \quad19^2=361\\ 10^2=100& \quad20^2= 400\hline\end(array)\]

fact 3.
How can you win with square roots?
$\bullet$ The sum of the sum of the square root is NOT equal to the square root of the sum of the sum of the square root, tobto \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] In this way, if you need to calculate, for example, $\sqrt(25)+\sqrt(49)$ , then it is your responsibility to know the value of $\sqrt(25)$ and $\sqrt(49)\ ), and then fold them. Otzhe, \[\sqrt(25)+\sqrt(49)=5+7=12\] Although the meaning of \(\sqrt a$ or $\sqrt b$ when adding $\sqrt a+\sqrt b$ is not known, then such a viraz is far from transforming and becoming so, like є. For example, for the sum $\sqrt 2+ \sqrt (49)$ we can know $\sqrt(49)$ - ce $7$ , but from $\sqrt 2$ it is impossible to convert , that $\sqrt 2+\sqrt(49)=\sqrt 2+7$. Dali tsey whistle, sorry, you can’t forgive$\bullet$ Twіr/private square root is equal to square root w/private, tobto \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(i)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (for the mind that offending parts of equivalence can sense)
Butt: $\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8$; $\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16$; $\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40$. $\bullet$ Koristuyuchis tsimi domineering, casually know the square root of the great numbers in a way of arranging them into multipliers.
Let's look at an example. We know $\sqrt(44100)$. Oskіlki $44100:100=441$, then $44100=100\cdot 441$. For the divisibility sign, the number $441$ is divided by $9$ (the sum of the digits is 9 and divided by 9), also, $441:9=49$ , then $441=9$ cdot 49).
In this rank, we took away: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
$\bullet$ It will be shown how to enter the numbers with the square root sign on the example $5\sqrt2$ (the abbreviations are written in the form $5\cdot \sqrt2$). Scope $5=\sqrt(25)$ , then \ Respectfully so, scho, for example,
1) $\sqrt2+3\sqrt2=4\sqrt2$ ,
2) $5\sqrt3-\sqrt3=4\sqrt3$
3) $\sqrt a+\sqrt a=2\sqrt a$ .

Why so? Let's explain from the butt 1). As you have already understood, it is impossible for us to transform the number (sqrt2). Obviously, $\sqrt2$ is the same number $a$. Vidpovidno, viraz $\sqrt2+3\sqrt2$ є nothing else, like $a+3a$ (one number $a$ plus three more of the same numbers $a$ ). And we know what is good for such numbers $a$, so $4\sqrt2$.

fact 4.
$\bullet$ It often seems "can't find the root" if you don't try to get the sign $\sqrt() \ $ of the root (radical) given the value of that number. For example, you can extract the root from the warehouse $16$ because $16=4^2$ , because $\sqrt(16)=4$ . And the axis to draw the root of the number $3$, so that you know $\sqrt3$, it is impossible, because there is no such number, so give $3$ in a square.
Such numbers (otherwise with such numbers) are irrational. For example, numbers $\sqrt3, \ 1+\sqrt2, \ \sqrt(15)$ etc. є irrational.
Also irrational є numbers $\pi$ (number "pi", approximately equal $3,14$ ), $e$ (this number is called the Euler number, approximately equal $2,7$ ) and etc.
$\bullet$ We give your respect to those, whether the number will be either rational or irrational. And at the same time, all rational and all irrational numbers establish the impersonal, which is called bezlіchchyu deysnyh (speech) numbers. It is denoted by the impersonal letter $\mathbb(R)$.
Henceforth, all numbers, as we know, are called speech numbers.

Fact 5.
$\bullet$ Modulus of a speech number $a$ is a number $|a|$ , which goes from the point $a$ to $0$ on the speech line. For example, $|3|$ and $|-3|$ equals 3, so that between points $3$ і $-3$ to $0$ however equal and equal $3 $.
$\bullet$ If $a$ is not a number, then $|a|=a$ .
Stock: $|5|=5$; $\qquad |\sqrt2|=\sqrt2$ . $\bullet$ If $a$ is a number, then $|a|=-a$ .
Stock: $|-5|=-(-5)=5$; $\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3$.
It seems that for negative numbers the module has a minus, and for positive numbers, as well as the number $0$, the module is left without change.
ALE such a rule is less suitable for numbers. If you don’t know $x$ under the sign of the module (because it’s not known otherwise), for example, $|x|$ , we don’t know about it, whether it’s positive, zero, or negative, then we don’t need to look at the module we can. І here cei vislіv is so i overflows: $|x|$ . $\bullet$ May have the same formula: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( mind you ) a\geqslant 0\] Even more often such a pardon is allowed: it seems that $\sqrt(a^2)$ and $(\sqrt a)^2$ are one and the same. Tse verno is less in that case, if $a$ is a positive number or zero. And from yakscho \ (a \) - a negative number, then not so. Dosit look at such an butt. Let's replace $a$ number $-1$. Then $\sqrt((-1)^2)=\sqrt(1)=1$ , and the axis viraz $(\sqrt (-1))^2$ is not necessary root place negative numbers!).
Therefore, I give your respect to those who $\sqrt(a^2)$ are not healthy $(\sqrt a)^2$! Stock: 1) $\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2$, because $-\sqrt2<0$ ;

$\phantom(00000)$ 2) $(\sqrt(2))^2=2$ . $\bullet$ Scales $\sqrt(a^2)=|a|$ , then \[\sqrt(a^(2n))=|a^n|\] (viraz $2n$ gives the guy a number)
So when the root is drawn from the number, which is known by the singing world, its foot changes into two.
Butt:
1) $\sqrt(4^6)=|4^3|=4^3=64$
2) $\sqrt((-25)^2)=|-25|=25$ 'yataєmo, that for the purpose of the root of such a buti cannot be: with us, when the root has been forgotten, it can come out positively, or a zero)
3) $\sqrt(x^(16))=|x^8|=x^8$

Fact 6.
How to equalize two square roots?
$\bullet$<\sqrt b\) , то $aButt:
1) \(\sqrt(50)$ and $6\sqrt2$ . For the cob, let's remake another viraz y $\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)$. In this rank, oskіlki \ (50<72\) , то и $\sqrt{50}<\sqrt{72}$ . Следовательно, $\sqrt{50}<6\sqrt2$ .
2) What number of integers does $\sqrt(50)$?
Skіlki $\sqrt(49)=7$ , $\sqrt(64)=8$ , and $49<50<64$ , то $7<\sqrt{50}<8$ , то есть число $\sqrt{50}$ находится между числами $7$ и $8$ .
3) Pair $\sqrt 2-1$ and $0.5$. Suppose that $\sqrt2-1>0.5$: \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((dodamo one to both parts))\\&\sqrt2>0,5+1 \ big| \ ^2 \quad\text((stars align squares))\\ &2>1,5^2\\ &2>2,25 \end(aligned)\] Bachimo, we got the wrong unevenness. Otzhe, our confession was nevirnim and $\sqrt 2-1<0,5$ .
Respectfully, adding the deaky number to both parts of the nervousness does not add the first sign. Multiplying/subdividing both parts of the unevenness on a positive number also does not add to the first sign, but multiplying/spending on a negative number changes the sign of unevenness to the protracted one!
It is possible to bring insults to parts of jealousy / nervousness into a square TILKI TODI, if insults are not part of nevid'emni. For example, in the unevenness of the front butt, you can make offense in the parts of the square, in the unevenness \ (-3<\sqrt2\) нельзя (убедитесь в этом сами)! $\bullet$ Next memory, what \[\begin(aligned) &\sqrt 2\approx 1,4\\ &\sqrt 3\approx 1,7 \end(aligned)\] Knowing the approximate value of these numbers will help you match the numbers! $\bullet$ In order to win the root (as it wins out) from such a great number, which is not in the table of squares, it is necessary to count on the back, between such “hundreds” one can be known, then - between such “tens”, and then select the remaining digit of that number. It will be shown how it works on the butt.
Take $\sqrt(28224)$. We know that $100^2=10\,000$, $200^2=40\,000$ and so on. Note that $28224$ is found between $10\,000$ and $40\,000$ . Also, $\sqrt(28224)$ is found between $100$ and $200$ .
Now it is significant that our number is between such “tens” (that is, for example, between (120) and (130)). It is also known from the square tables that $11^2=121$ , $12^2=144$ and so on, then$110^2=12100$ , $120^2=14400 ) $ , $130^2=16900$ , $140^2=19600$ , $150^2=22500$ , $160^2=25600$ , $170^2= 28900 $. In this order, we believe that $28224$ is found between $160^2$ and $170^2$ . Also, the number $\sqrt(28224)$ is between $160$ and $170$.
Try to figure out the remaining number. Let's guess, what are single-digit numbers when squared to give on the end \ (4 \) ? Tse $2^2$ and $8^2$. Later, $\sqrt(28224)$ will end either by 2 or by 8. Let's reverse it. We know $162^2$ and $168^2$:
$162^2=162\cdot 162=26224$
$168^2=168\cdot 168=28224$ .
Also, $\sqrt(28224) = 168$. Veil!

In order to be able to verse EDI s of mathematics, we must first learn the theoretical material, which is known from numerical theorems, formulas, and algorithms. At first glance, you can see what is easy to finish. However, to know the dzherelo, in which the theory for EDI s of mathematics was easily and wisely developed for students with any level of training, - it’s more difficult to finish the task. Shkіlnі podruchniki impossibly zavzhd trimati under hand. And to know the basic formulas for EDI in mathematics is not easy to learn from the Internet.

Why is it so important to teach the theory of mathematics, not only for those who build ED?

Because I expand my horizons. The introduction of theoretical material from mathematics is dear to everyone who wants to take into account the wide range of food, which is familiar to the world of knowledge. Everything in nature is in order and can read logic. The very same is seen in science, through the yak you can understand the world.
Because it develops intellect. Vivchayuchi dovіdkovі materials for ЄДІz mathematics, as well as viruyuyuchi raznomanіtnі zavdannya, people learn to think logically and rozmіrkovuvati, competently and clearly formulate ideas. The new one has the ability to analyze, zagalnyuvati, robiti visnovki.

We encourage you to especially evaluate all the advantages of our approach to the systematization and presentation of the initial materials.

Step formulas vikoristovuyut in the process of quickness and forgiveness of folding viruses, in the virishennі rіvnіan and irritability.

Number cє n-th step of the number a if:

Operations in steps.

1. Multiplying the steps with the same basis, their indicators are added up:

a ma n = a m + n.

2. At the rozpodіlі stаіnіv z the same basis їх pokanika vіdnіmayutsya:

3. Steps of practice of the 2nd chi greater number of multipliers in more advanced steps of these sp_multipliers:

(abc…) n = a n b n c n …

4. The steps of the fraction are more advanced in the introduction of the steps of a given one:

(a/b) n = n/b n .

5. The stars of the steps at the feet, the indicators of the steps are multiplied:

(am) n = a m n .

The skin is shown formula virna u straight ahead zliva to the right and navpak.

For example. (2 3 5/15)² = 2² 3² 5²/15² = 900/225 = 4.

Operations with roots.

1. The root of the creation of many spivmulniki in dobrivnyu dobutku root of these spivmulniki:

2. Root from the root of the root of the root:

3. When the root is added to the rіven, add the zvedi to the whole rіven the root number:

4. How to increase the root steps in n once i at the same hour call in n-th step of the root number, then the value of the root does not change:

5. How to change the root steps in n once and at the same time, pull up the roots n-th step from the root number, then the value of the root does not change:

Step out of a negative indicator. The step of the same number with a non-positive (qіlim) indicator is assigned as one, divided by the step of the same number with the indicator, which is equal to the absolute value of the non-positive indicator:

Formula a m:a n = a m - n you can win not only for m> n, ale i at m< n.

For example. a4: a 7 = a 4 - 7 = a -3.

Schob formula a m:a n = a m - n became fair at m=n, the presence of the zero step is required.

Step out of the zero indicator. A step, be it a number, which is not equal to zero, with a zero indicator of a good one.

For example. 2 0 = 1,(-5) 0 = 1,(-3/5) 0 = 1.

Step out of the shotgun display. Schob to call a day number but at the feet m/n it is necessary to win the root n oh world z m th step of th number but.