Links of the square root to the feet. Reproduction of roots: basic rules. Rozpodil pіdroenih virazіv

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:

1. 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.
2. 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 number 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, it is possible to recognize, most of the unprepared scientists in the back do not work out 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 grade 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 viraz 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:

1. Get on the wall to state that mathematics is a stupid science, that there are rules, but it’s still inaccurate;
2. 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 minus 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:

1. 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 you need speed (for example, there are two minuses).
2. 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. 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:

1. 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.
2. 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?

Apparently, the sign of the root is the square root of the song number. Prote the sign of the root means as an algebraic work, and y zastosovyatsya in woodworking production - for visual expansions.

If you want to know, how to multiply the root "z" or "without" multiples, then this article is for you. We have a look at the method of multiplying the root:

• without multiples;
• with multipliers;
• with various indications.

Root multiplication method without multipliers

Algorithm diy:

Perekonatisya, what are the roots of the same signs (step). Guess what steps are written evil over the sign of the root. There is no sign of a step, tse, what is the square root, tobto. with step 2, and yogo can be multiplied by another root with step 2.

Butt

Stock 1: 18×2 = ?

Stock 2: 10×5 = ?

Butt

Stock 1: 18×2 = 36

Stock 2: 10x5 = 50

Stock 3: 3 3 × 9 3 = 27 3

Forgive the root of the Viraz. If we multiply the root one by one, we can forgive the subtraction of the roots of the viraz to the creation of the number (or viraz) by the full square or cube:

Butt

Example 1: 36 \u003d 6. 36 is the square root of six (6 × 6 \u003d 36).

Stock 2: 50 = (25 × 2) = (5 × 5) × 2 = 5 2 . The number 50 is laid out on tvir 25 and 2. Root z 25 - 5 to that, 5 z-pod sign of the root і is simply viraz.

Stock 3: 273 = 3. The cube root of 27 is old 3: 3 × 3 × 3 = 27.

The method of multiplying indications with multiples

Algorithm diy:

Multiply multiples. A multiplier is a number, like standing in front of the root sign. At the time of the presence of a multiplier of wines, for locks, it is respected by loneliness. Further, it is necessary to multiply the factors:

Butt

Stock 1: 3 2 × 10 = 3? 3 x 1 = 3

Stock 2: 4 3 × 3 6 = 12? 4x3=12

Multiply the numbers under the sign of the root. Just as you have multiplied the multipliers, boldly multiply the numbers to stand under the sign of the root:

Butt

Stock 1: 3 2 × 10 = 3 (2 × 10) = 3 20

Stock 2: 4 3 × 3 6 = 12 (3 × 6) = 12 18

Forgive the root of the viraz. We gave the following values, which stand under the sign of the root - it is necessary to blame the corresponding numbers for the sign of the root. Then it is necessary to multiply the numbers and multipliers, as if to stand in front of the root sign:

Butt

Stock 1: 3 20 = 3 (4 × 5) = 3 (2 × 2) × 5 = (3 × 2) 5 = 6 5

Stock 2: 12 18 = 12 (9 × 2) = 12 (3 × 3) × 2 = (12 × 3) 2 = 36 2

The method of multiplying the root with different indications

Algorithm diy:

Find the least significant multiple (LCM) of indications. The least significant multiple is the least number that can be divided into offensive displays.

Butt

It is necessary to know the NOC of indications for offensive virus:

Indicators are rising 3 and 2 . For tsikh two numbers, the smallest major multiple of є is the number 6 (it can be divided without excess і by 3 і by 2). For multiplying the root, an indication is needed 6.

Write down skin manifestations from a new indicator:

Know the numbers, it is necessary to multiply the showings by yaki, in order to take the NOC.

For virus 5 3 it is necessary to multiply 3 by 2 to subtract 6 . And in case of virus 2 2 - in the same way, it is necessary to multiply by 3 to subtract 6.

Call the number, as if to stand under the sign of the root, in an equal number, as it was found in the front crochet. For the first virase 5, it is necessary to bring it up to step 2, and the other - 2 to step 3:

2 → 5 6 = 5 2 6 3 → 2 6 = 2 3 6

Call at the feet of the viraz and write down the result under the sign of the root:

5 2 6 = (5 × 5) 6 = 25 6 2 3 6 = (2 × 2 × 2) 6 = 8 6

Multiply the numbers under the roots:

(8×25) 6

Write result:

(8 × 25) 6 = 200 6

If possible, it is necessary to ask for forgiveness, but at times it will not be asked.

How did you remember the pardon in the text, be kind, see it and press Ctrl + Enter

The presence of square roots in the expression makes the process of solving the problem, prote the rules, for the help of such a robot with fractions, it becomes much simpler.

The only thing that is necessary for the whole hour of remembrance- sub-roots of virazi are subdivided into sub-roots of virazi, and multipliers to multipliers. In the process of rozpodіlu square roots we will simply drіb. So, we guess, what a root can be at the bannerman.

Method 1. Rosepodil root virus

Algorithm diy:

Write down drib

If you don’t have any idea when looking at a fraction, it’s necessary to write it down like that, it’s easier for him to understand the principle of splitting square roots.

butt 1

144 ÷ 36

Victory one root sign

Likewise, in the numeral book, and the banner of the presence of square roots, it is necessary to write down their sub-roots under one sign of the root, so that the process of ending is simpler.

It’s guessing that under the root of the virase (or number) is under the sign of the root.

butt 2

144 36 . Cey viraz should be written like this: 144 36

Expand Virazi root roots

Just divide one virase into another, and write the result under the sign of the root.

butt 3

144 36 = 4

Forgive the root of the virus (as it is necessary)

For example, one of the multiples is a perfect square, ask for such a viraz.

Let's guess that the square is the number, like the square of the third whole number.

butt 4

4 is a complete square, so 2×2 = 4 . Whose next:

4 = 2×2 = 2 . Tom 144 36 = 4 = 2.

Method 2. Multiplication of root virase

Algorithm diy:

Write down drib

Rewrite viraz at the sight of a shot (as it is presented like this). It makes the process of splitting viraziones with square roots much easier, especially when splitting into multiples.

butt 5

8 ÷ 36 , rewritten as 8 36

Spread on multipliers of skins from root viruses

The number under the root is divided into multipliers, as if it were another number, only the multipliers should be written under the sign of the root.

butt 6

8 36 = 2 x 2 x 2 6 x 6

Forgive the number and the banner of the fraction

For which next, add the multipliers to the sign of the root, which are equal squares. In this rank, the multiplier of the sub-root virazu will become a multiplier in front of the root sign.

butt 7

2 2 6 6 × 6 2 × 2 × 2

Rationalize the banner (take root)

Mathematics has rules that make it impossible to root a bannerman - a sign of a filthy tone, tobto. can't. If there is a square root in the banner, then indulge in yoga.

Multiply the number book and the banner by the square root, which you need to ask.

butt 8

In case of virazi 6 2 3 it is necessary to multiply the number and banner by 3 so that Yogo will appear in the banner:

6 2 3 × 3 3 = 6 2 × 3 3 × 3 = 6 6 9 = 6 6 3

Forgive otrimaniy viraz (as necessary)

Like in the number book and the banner of the presence of numbers, like it is possible and necessary to be quick. Ask like this virazi, as if it were something else.

butt 9

2 6 ask to 1 3; in this order 2 2 6 ask to 1 2 3 \u003d 2 3

Method 3. Subdivision of square root with multipliers

Algorithm diy:

Forgive multipliers

Let's guess that the multipliers are numbers, which should be in front of the root sign. For the forgiveness of multiples, it is necessary to divide or speed them up. Don't chip the root of the viraz!

butt 10

4 32 6 16 . Spochatku quickly 46: divisible by 2 - numeral, і banner: 46 \u003d 23.

Forgive the square root

Like a numeral, it is divided into a banner, divide. Like, ask for the root words, like and be-yakі іnshі.

butt 11

32 to be divided on the 16th volume: 32 16 = 2

Multiply the simple multipliers by the simple root

Remember the rule: do not deprive the flagman of the root. To this we simply multiply the number and the banner of the root.

butt 12

2 3 × 2 = 2 2 3

Rationalize the bannerman (take root at the bannerman)

butt 13

4 3 2 7 . Next, multiply the number book and the banner by 7 so that the root will be spared from the banner.

4 3 7 × 7 7 = 4 3 × 7 7 × 7 = 4 21 49 = 4 21 7

Method 4

Algorithm diy:

Significantly, chi binomial (bіnom) at the bannerman

Guessing that the binary is a virase, which includes 2 monomials. Such a method can only be used in vipads, if there is a binary with a square root in the standard.

butt 14

1 5 + 2 - the banner has a presence bіnom, oskіlki є two monomials.

Know viraz, bind with bean

Let's guess that there were binomials and binomials with these same monomials, but also with opposite signs. In order to forgive the viraz and spare the root in the banner, multiply the pov'yazan bіnom.

butt 15

5 + 2 і 5 - 2 - received bіnom.

Multiply the numeral and the banner by the binary, which is a bandage for the banner

Such an option will help to save the root in the banner, shards of additional binomials in the other squares of the skin term of the binomials: (a - b) (a + b) \u003d a 2 - b 2

butt 16

1 5 + 2 = 1 (5 - 2) (5 - 2) (5 + 2) = 5 - 2 (5 2 - (2) 2 = 5 - 2 25 - 2 = 5 - 2 23 .

On the next track: 1 5 + 2 \u003d 5 - 2 23.

Please:

1. If you work with the square roots of mixed numbers, then convert them to wrong numbers.
2. Vіdmіnnіst dоdаvannya i vіdnіmannyа vіd rozpodіlu - podkorіnі vyslovlyuvannya razі rozpodіlu is not recommended to be asked (for the rahunok of the new squares).
3. Nikoli (!) Do not deprive the flagman of the root.
4. Everyday tens of fractions or mixed before the root - it is necessary to transform them into the greatest fraction, and then forgive.
5. Does the bannerman have a sum of chi ration of two monomers? Multiply such a binomial by binding to you binomial and leaving the root in the banner.

How did you remember the pardon in the text, be kind, see it and press Ctrl + Enter

Why is it necessary to build a foldable rozrahunka, but an electronic calculation outbuilding did not appear at hand? Speed ​​up with the online program - the root calculator. Vaughn help:

• know the square or cubic root of given numbers;
• vikonati mathematical action with shot steps.

Number of characters after Komi:

√

How to calculate the square root manually - by the selection method you know the exact values. Let's look at it like a robiti.

What is the square root

Korin n step of natural quantity a- Number, n steps of someone else a(sub-root number). The root is denoted by the symbol √. Yogo is called a radical.

The skin is mathematical for the protidium: adding → vіdnіmannya, multiplying → podіl, zvedennya in the foot → forcing the root.

The square root of a number a there will be a number, the square of some kind a. Why do you need food, how can you calculate the root of the number? It is necessary to choose a number, as if another world had a more valuable value under the root.

Sound 2 do not write above the root sign. Shards of the smallest step, obviously not a number, then it is possible to take away the indicator 2. It is clear: to calculate the square root of 16, it is necessary to know the number, when you add it to another step, enter 16.

We carry out rozrahunki manually

Calculation by the method of spreading on simple multipliers is counted in two ways, depending on the fact that the root number is:

1. Tsіle, yak you can lay out on square multipliers and take the exact answer.

Square numbers - numbers from which you can take the root without excess. And multipliers are numbers, which, when multiplied, give the same number.

For example:

25, 36, 49 - square numbers, shards:

Come out, that square multipliers are multipliers, like square numbers.

Vіzmemo 784 and vityagnemo z new root.

Let's decompose the number into square factors. The number 784 is a multiple of 4, so the first square factor is 4 x 4 = 16. If we divide 784 by 16, we can take 49 - the whole square number is 7 x 7 = 16.
Zastosuєmo rule Let's take the root of the skin square multiplier, we'll multiply the results and we'll take it.	Vidpovid.

2. Inappropriate. Yoga cannot be divided into square multipliers.

Such butts are used more often, lower in whole numbers. Your decision will not be exact, we want it. We will be closer and closer. Forgive the task in addition to laying out the root number by the square factor, that number, from which it is impossible to extract the square root.

Let's spread the number 252 to the square and the greatest multiplier.
Estimate the value of the root. For this we select two square numbers, which are to stand in front and behind the root number in the digital line.	The root number is 7. The nearest square number will be 8, and less than 4. between 2 and 4.
Estimated value	Naimovіrnіshe √7 is closer to 2. It is chosen in such a rank, so that with a multiplied number, 7 appeared on itself. 2.7 x 2.7 = 7.2. Not suitable, scaling 7.2>7, less than 2.6 x 2.6 = 6.76. Zalishaemo, adzhe 6,76-7.
Calculating the root

How to virahuvati the root of a folding number? Tezh by the method of estimating the value of the root.

When rozpodіlі in stovpchik go out as accurately as possible vіdpovіd pіd hour of root extrusion.

Take the arch of paper and re-arm it so that the vertical line is in the middle, and the horizontal bula is from the right side and below the spadix.
Break the root number into sprats of numbers. Divide decimals like this: - whole part on the right side; - The number after komi zliva to the right.	Stock: 3459842.825694 → 3 45 98 42, 82 56 94 795,28 → 7 95, 28 It is allowed that an unpaired number is overwritten.
For the first number (or a bet), we choose the most n. Yogo square can be smaller or more important than the first number (bet numbers). Vyymіt іz th number korіn - √n. Write down the subtraction result to the right-handed beast, and the square of the number - to the right-handed bottom. We have persha 7. The nearest square number is 4. It is less than 7, and 4 =
See the knowledge of the square of the number n from the first number (bet). Write down the result pid 7. And the upper number of the right-hander is subtracted and written down by the right-hander 4_х_=_. Note: numbers may be the same.
We select the number for the slashes. For whom to know such a number, so that the omission of twir is not greater, but more dignified by the current number of evil. Our vpadka, tse 8.
Write down the found number at the upper right corner. This is another number from the root of the joke. I’ll come up with a couple of numbers and write down the amount of money taken away from the money.
Vіdnіmіt otrimaniy on the right tvіr іz lіvoruch. Subvoyuyemo number, as if roztashovane right-handed eel and write down viraz іz dashes.
We bring to retail, what happened, a couple of numbers. As if the numbers of the shot part, tobto roztashovani behind the coma, then at the upper right fold there was the remaining digit of the square root, just to joke, put to whom. Let’s fill in the dashes at the right-handed virazi, pick the number in such a way that the truncated tvir will be smaller or more expensive viraz zliva.
If you need more than the number of signs after the Komi, then add the number of current figures of Zliva and repeat dії: add the number of Zliva, subtract the number from the upper right fold, write it down with dashes, select the multipliers for the new one and so on.

What do you think, how many hours do you spend on such rosrahunki? Difficult, long, confused. Then why not ask yourself the manager? Hurry up with our program, as if to help you grow swedes and exactly roses.

Algorithm diy

1. Enter the number of signs after Komi.

2. Specify the step of the root (for example, the larger one for 2).

3. Enter the number for which you plan to root.

4. Press the "Submit" button.

We will forgive the calculation of the most complicated mathematical diy with an online calculator!

The study of the quadrant root of a warehouse is not a single operation, as it can be carried out with this mathematical phenomenon. So, just like the primary numbers, the square roots add up and see.

Folding rules and vіdnіmannya square root

Appointment 1

So dії, like adding and seeing the square root, you can only think of the same root virase.

butt 1

You can fold or see virazi 2 3 that 6 3, but not 5 6 і 9 4 . If it is possible to ask the viraz and bring it to the root with the same root number, ask, and then fold or see.

Dії z korіnnyam: the basics

butt 2

6 50 - 2 8 + 5 12

Algorithm dії:

1. Sorry pіdkorene viraz. For which it is necessary to spread the root of the virus into 2 multipliers, one of them is a square number (the number from which the square root is drawn, for example, 25 or 9).
2. Let's sweat the root of the square number that write down the otrimane value before the root sign. We respect your respect that another multiplier is entered under the sign of the root.
3. After the process of parsing, it is necessary to add a root with the same root virase - only a few of them can be added and seen.
4. At the root with the same root virases, it is necessary to add or add multipliers, as if to stand in front of the root sign. The root of the virus is left without changes. You can’t add up chi vodnіmati pіdkorіnі numbers!

Porada 1

If you have a butt with a great number of the same sub-root verses, then add such verses with single, subtractive and triple lines, in order to facilitate the process of calculation.

butt 3

Let's try to break the whole butt:

6 50 = 6 (25×2) = (6×5) 2 = 30 2 . For the cob, it is necessary to spread 50 by 2 multipliers 25 and 2, then we will take the root of 25, which is 5, and 5 will blame the root of the root. Then it is necessary to multiply 5 by 6 (multiplier at the root) and gain 30 2 .

2 8 = 2 (4×2) = (2×2) 2 = 4 2 . First, it is necessary to spread 8 by 2 multipliers: 4 and 2. Let’s take the root out of 4, which is a good 2, and 2 to blame the root of the root. If necessary, multiply 2 by 2 (multiplying the root) and take 4 2.

5 12 = 5 (4×3) = (5×2) 3 = 10 3 . It is necessary to spread the back of the hand 12 by 2 multipliers: 4 and 3. Let's win from the 4th root, which is a good 2, and blame the yogo root. If necessary, multiply 2 by 5 (multiplier at the root) and subtract 103.

Asking result: 30 2 - 4 2 + 10 3

30 2 - 4 2 + 10 3 = (30 - 4) 2 + 10 3 = 26 2 + 10 3 .

At the result, they succumbed, but some of the same roots of language can be avenged at this butt. And now we will practice on other stocks.

butt 4

• Let's just say (45) . We spread 45 into multipliers: (45) = (9 × 5);
• Vinosimo 3 z-pіd roots (9 \u003d 3): 45 \u003d 3 5;
• We add the multipliers of the root: 3 5 + 4 5 \u003d 7 5.

butt 5

6 40 - 3 10 + 5:

• Let's just say 6 40 . We put 40 into multipliers: 640 \u003d 6 (4 × 10);
• Vinosimo 2 z-pіd roots (4 = 2): 6 40 = 6 (4 × 10) = (6 × 2) 10;
• We multiply the multipliers to stand in front of the roots: 12 10;
• It is written down in a simple way: 12 10 - 3 10 + 5;
• The first two terms can have the same root numbers, we can see them: (12 - 3) 10 \u003d 9 10 + 5.

butt 6

As we know, it’s impossible to justify the root of the number, so we joke in the application of the terms with the same root numbers, conduct mathematical divisions (add, see, etc.) and record the result:

(9 - 4) 5 - 2 3 = 5 5 - 2 3 .

Please:

• Before that, how to fold or see, it is necessary to obov'yazkovo ask (as it is possible) rooted virazi.
• Putting that visible root with different rooted virazes is drastically defended.
• Do not follow the sum of the numbers to see the number of the roots: 3 + (2 x) 1/2.
• When counting with fractions, it is necessary to know the number, so that it should be divided into a leather banner, then bring the fractions to the full banner, then add up the numbers, and leave the banners without changes.

How did you remember the pardon in the text, be kind, see it and press Ctrl + Enter