The square root of a number X called a number A, which in the process of multiplying itself by itself ( A*A) can give a number X.
Those. A * A = A 2 = X, and √X = A.

Over square roots ( √x), as with other numbers, you can perform arithmetic operations such as subtraction and addition. To subtract and add roots, they must be connected using signs corresponding to these actions (for example √x- √y ).
And then bring the roots to their simplest form - if there are similar ones between them, you need to make a cast. It consists in the fact that the coefficients of similar terms are taken with the signs of the corresponding terms, then they are enclosed in brackets and the common root is displayed outside the multiplier brackets. The coefficient that we have obtained is simplified according to the usual rules.

Step 1. Extracting square roots

First, to add square roots, you first need to extract these roots. This can be done if the numbers under the root sign are perfect squares. For example, take the given expression √4 + √9 . First number 4 is the square of the number 2 . Second number 9 is the square of the number 3 . Thus, the following equality can be obtained: √4 + √9 = 2 + 3 = 5 .
Everything, the example is solved. But it doesn't always happen that way.

Step 2. Taking out the multiplier of a number from under the root

If there are no full squares under the root sign, you can try to take the multiplier of the number out from under the root sign. For example, take the expression √24 + √54 .

Let's factorize the numbers:
24 = 2 * 2 * 2 * 3 ,
54 = 2 * 3 * 3 * 3 .

In list 24 we have a multiplier 4 , it can be taken out from under the square root sign. In list 54 we have a multiplier 9 .

We get the equality:
√24 + √54 = √(4 * 6) + √(9 * 6) = 2 * √6 + 3 * √6 = 5 * √6 .

Considering this example, we get the removal of the factor from under the root sign, thereby simplifying the given expression.

Step 3. Reducing the denominator

Consider the following situation: the sum of two square roots is the denominator of a fraction, for example, A / (√a + √b).
Now we are faced with the task of "getting rid of the irrationality in the denominator."
Let's use the following method: multiply the numerator and denominator of the fraction by the expression √a - √b.

We now get the abbreviated multiplication formula in the denominator:
(√a + √b) * (√a - √b) = a - b.

Similarly, if the denominator contains the difference of the roots: √a - √b, the numerator and denominator of the fraction are multiplied by the expression √a + √b.

Let's take a fraction as an example:
4 / (√3 + √5) = 4 * (√3 - √5) / ((√3 + √5) * (√3 - √5)) = 4 * (√3 - √5) / (-2) = 2 * (√5 - √3) .

An example of complex denominator reduction

Now we will consider a rather complicated example of getting rid of irrationality in the denominator.

Let's take a fraction as an example: 12 / (√2 + √3 + √5) .
You need to take its numerator and denominator and multiply by the expression √2 + √3 - √5 .

We get:

12 / (√2 + √3 + √5) = 12 * (√2 + √3 - √5) / (2 * √6) = 2 * √3 + 3 * √2 - √30.

Step 4. Calculate the approximate value on the calculator

If you only need an approximate value, this can be done on a calculator by calculating the value of square roots. Separately, for each number, the value is calculated and recorded with the required accuracy, which is determined by the number of decimal places. Further, all the required operations are performed, as with ordinary numbers.

Estimated Calculation Example

It is necessary to calculate the approximate value of this expression √7 + √5 .

As a result, we get:

√7 + √5 ≈ 2,65 + 2,24 = 4,89 .

Please note: under no circumstances should square roots be added as prime numbers, this is completely unacceptable. That is, if you add the square root of five and three, we cannot get the square root of eight.

Useful advice: if you decide to factorize a number, in order to derive a square from under the root sign, you need to do a reverse check, that is, multiply all the factors that resulted from the calculations, and the final result of this mathematical calculation should be the number we were originally given.

Fact 1.
\(\bullet\) Take some non-negative number \(a\) (ie \(a\geqslant 0\) ). Then (arithmetic) square root from the number \(a\) such a non-negative number \(b\) is called, when squaring it we get the number \(a\) : \[\sqrt a=b\quad \text(same as )\quad a=b^2\] It follows from the definition that \(a\geqslant 0, b\geqslant 0\). These restrictions are an important condition for the existence of a square root and should be remembered!
Recall that any number when squared gives a non-negative result. That is, \(100^2=10000\geqslant 0\) and \((-100)^2=10000\geqslant 0\) .
\(\bullet\) What is \(\sqrt(25)\) ? We know that \(5^2=25\) and \((-5)^2=25\) . Since by definition we have to find a non-negative number, \(-5\) is not suitable, hence \(\sqrt(25)=5\) (since \(25=5^2\) ).
Finding the value \(\sqrt a\) is called taking the square root of the number \(a\) , and the number \(a\) is called the root expression.
\(\bullet\) Based on the definition, the expressions \(\sqrt(-25)\) , \(\sqrt(-4)\) , etc. don't make sense.

Fact 2.
For quick calculations, it will be useful to learn 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.
What can be done with square roots?
\(\bullet\) The sum or difference of square roots is NOT EQUAL to the square root of the sum or difference, i.e. \[\sqrt a\pm\sqrt b\ne \sqrt(a\pm b)\] Thus, if you need to calculate, for example, \(\sqrt(25)+\sqrt(49)\) , then initially you must find the values ​​\(\sqrt(25)\) and \(\sqrt(49)\ ) and then add them up. Consequently, \[\sqrt(25)+\sqrt(49)=5+7=12\] If the values ​​\(\sqrt a\) or \(\sqrt b\) cannot be found when adding \(\sqrt a+\sqrt b\), then such an expression is not further converted and remains as it is. For example, in the sum \(\sqrt 2+ \sqrt (49)\) we can find \(\sqrt(49)\) - this is \(7\) , but \(\sqrt 2\) cannot be converted in any way, that's why \(\sqrt 2+\sqrt(49)=\sqrt 2+7\). Further, this expression, unfortunately, cannot be simplified in any way.\(\bullet\) The product/quotient of square roots is equal to the square root of the product/quotient, i.e. \[\sqrt a\cdot \sqrt b=\sqrt(ab)\quad \text(s)\quad \sqrt a:\sqrt b=\sqrt(a:b)\] (provided that both parts of the equalities make sense)
Example: \(\sqrt(32)\cdot \sqrt 2=\sqrt(32\cdot 2)=\sqrt(64)=8\); \(\sqrt(768):\sqrt3=\sqrt(768:3)=\sqrt(256)=16\); \(\sqrt((-25)\cdot (-64))=\sqrt(25\cdot 64)=\sqrt(25)\cdot \sqrt(64)= 5\cdot 8=40\). \(\bullet\) Using these properties, it is convenient to find the square roots of large numbers by factoring them.
Consider an example. Find \(\sqrt(44100)\) . Since \(44100:100=441\) , then \(44100=100\cdot 441\) . According to the criterion of divisibility, the number \(441\) is divisible by \(9\) (since the sum of its digits is 9 and is divisible by 9), therefore, \(441:9=49\) , that is, \(441=9\ cdot 49\) .
Thus, we got: \[\sqrt(44100)=\sqrt(9\cdot 49\cdot 100)= \sqrt9\cdot \sqrt(49)\cdot \sqrt(100)=3\cdot 7\cdot 10=210\] Let's look at another example: \[\sqrt(\dfrac(32\cdot 294)(27))= \sqrt(\dfrac(16\cdot 2\cdot 3\cdot 49\cdot 2)(9\cdot 3))= \sqrt( \ dfrac(16\cdot4\cdot49)(9))=\dfrac(\sqrt(16)\cdot \sqrt4 \cdot \sqrt(49))(\sqrt9)=\dfrac(4\cdot 2\cdot 7)3 =\dfrac(56)3\]
\(\bullet\) Let's show how to enter numbers under the square root sign using the example of the expression \(5\sqrt2\) (short for the expression \(5\cdot \sqrt2\) ). Since \(5=\sqrt(25)\) , then \ Note also that, for example,
1) \(\sqrt2+3\sqrt2=4\sqrt2\) ,
2) \(5\sqrt3-\sqrt3=4\sqrt3\)
3) \(\sqrt a+\sqrt a=2\sqrt a\) .

Why is that? Let's explain with example 1). As you already understood, we cannot somehow convert the number \(\sqrt2\) . Imagine that \(\sqrt2\) is some number \(a\) . Accordingly, the expression \(\sqrt2+3\sqrt2\) is nothing but \(a+3a\) (one number \(a\) plus three more of the same numbers \(a\) ). And we know that this is equal to four such numbers \(a\) , that is, \(4\sqrt2\) .

Fact 4.
\(\bullet\) It is often said “cannot extract the root” when it is not possible to get rid of the sign \(\sqrt () \ \) of the root (radical) when finding the value of some number. For example, you can root the number \(16\) because \(16=4^2\) , so \(\sqrt(16)=4\) . But to extract the root from the number \(3\) , that is, to find \(\sqrt3\) , it is impossible, because there is no such number that squared will give \(3\) .
Such numbers (or expressions with such numbers) are irrational. For example, numbers \(\sqrt3, \ 1+\sqrt2, \ \sqrt(15)\) etc. are irrational.
Also irrational are the numbers \(\pi\) (the number “pi”, approximately equal to \(3,14\) ), \(e\) (this number is called the Euler number, approximately equal to \(2,7\) ) etc.
\(\bullet\) Please note that any number will be either rational or irrational. And together all rational and all irrational numbers form a set called set of real (real) numbers. This set is denoted by the letter \(\mathbb(R)\) .
This means that all the numbers that we currently know are called real numbers.

Fact 5.
\(\bullet\) Modulus of a real number \(a\) is a non-negative number \(|a|\) equal to the distance from the point \(a\) to \(0\) on the real line. For example, \(|3|\) and \(|-3|\) are equal to 3, since the distances from the points \(3\) and \(-3\) to \(0\) are the same and equal to \(3 \) .
\(\bullet\) If \(a\) is a non-negative number, then \(|a|=a\) .
Example: \(|5|=5\) ; \(\qquad |\sqrt2|=\sqrt2\) . \(\bullet\) If \(a\) is a negative number, then \(|a|=-a\) .
Example: \(|-5|=-(-5)=5\) ; \(\qquad |-\sqrt3|=-(-\sqrt3)=\sqrt3\).
They say that for negative numbers, the module “eats” the minus, and positive numbers, as well as the number \(0\) , the module leaves unchanged.
BUT this rule only applies to numbers. If you have an unknown \(x\) (or some other unknown) under the module sign, for example, \(|x|\) , about which we do not know whether it is positive, equal to zero or negative, then get rid of the module we can not. In this case, this expression remains so: \(|x|\) . \(\bullet\) The following formulas hold: \[(\large(\sqrt(a^2)=|a|))\] \[(\large((\sqrt(a))^2=a)), \text( provided ) a\geqslant 0\] The following mistake is often made: they say that \(\sqrt(a^2)\) and \((\sqrt a)^2\) are the same thing. This is true only when \(a\) is a positive number or zero. But if \(a\) is a negative number, then this is not true. It suffices to consider such an example. Let's take the number \(-1\) instead of \(a\). Then \(\sqrt((-1)^2)=\sqrt(1)=1\) , but the expression \((\sqrt (-1))^2\) does not exist at all (because it is impossible under the root sign put negative numbers in!).
Therefore, we draw your attention to the fact that \(\sqrt(a^2)\) is not equal to \((\sqrt a)^2\) ! Example: 1) \(\sqrt(\left(-\sqrt2\right)^2)=|-\sqrt2|=\sqrt2\), because \(-\sqrt2<0\) ;

\(\phantom(00000)\) 2) \((\sqrt(2))^2=2\) . \(\bullet\) Since \(\sqrt(a^2)=|a|\) , then \[\sqrt(a^(2n))=|a^n|\] (the expression \(2n\) denotes an even number)
That is, when extracting the root from a number that is in some degree, this degree is halved.
Example:
1) \(\sqrt(4^6)=|4^3|=4^3=64\)
2) \(\sqrt((-25)^2)=|-25|=25\) (note that if the module is not set, then it turns out that the root of the number is equal to \(-25\) ; but we remember , which, by definition of the root, this cannot be: when extracting the root, we should always get a positive number or zero)
3) \(\sqrt(x^(16))=|x^8|=x^8\) (since any number to an even power is non-negative)

Fact 6.
How to compare two square roots?
\(\bullet\) True for square roots: if \(\sqrt a<\sqrt b\) , то \(aExample:
1) compare \(\sqrt(50)\) and \(6\sqrt2\) . First, we transform the second expression into \(\sqrt(36)\cdot \sqrt2=\sqrt(36\cdot 2)=\sqrt(72)\). Thus, since \(50<72\) , то и \(\sqrt{50}<\sqrt{72}\) . Следовательно, \(\sqrt{50}<6\sqrt2\) .
2) Between which integers is \(\sqrt(50)\) ?
Since \(\sqrt(49)=7\) , \(\sqrt(64)=8\) , and \(49<50<64\) , то \(7<\sqrt{50}<8\) , то есть число \(\sqrt{50}\) находится между числами \(7\) и \(8\) .
3) Compare \(\sqrt 2-1\) and \(0,5\) . Suppose \(\sqrt2-1>0.5\) : \[\begin(aligned) &\sqrt 2-1>0.5 \ \big| +1\quad \text((add one to both sides))\\ &\sqrt2>0.5+1 \ \big| \ ^2 \quad\text((square both parts))\\ &2>1,5^2\\ &2>2,25 \end(aligned)\] We see that we have obtained an incorrect inequality. Therefore, our assumption was wrong and \(\sqrt 2-1<0,5\) .
Note that adding a certain number to both sides of the inequality does not affect its sign. Multiplying/dividing both parts of the inequality by a positive number also does not affect its sign, but multiplying/dividing by a negative number reverses the sign of the inequality!
Both sides of an equation/inequality can be squared ONLY IF both sides are non-negative. For example, in the inequality from the previous example, you can square both sides, in the inequality \(-3<\sqrt2\) нельзя (убедитесь в этом сами)! \(\bullet\) Note that \[\begin(aligned) &\sqrt 2\approx 1,4\\ &\sqrt 3\approx 1,7 \end(aligned)\] Knowing the approximate meaning of these numbers will help you when comparing numbers! \(\bullet\) In order to extract the root (if it is extracted) from some large number that is not in the table of squares, you must first determine between which “hundreds” it is, then between which “tens”, and then determine the last digit of this number. Let's show how it works with an example.
Take \(\sqrt(28224)\) . We know that \(100^2=10\,000\) , \(200^2=40\,000\) and so on. Note that \(28224\) is between \(10\,000\) and \(40\,000\) . Therefore, \(\sqrt(28224)\) is between \(100\) and \(200\) .
Now let's determine between which “tens” our number is (that is, for example, between \(120\) and \(130\) ). We also know from the table of squares that \(11^2=121\) , \(12^2=144\) etc., then \(110^2=12100\) , \(120^2=14400 \) , \(130^2=16900\) , \(140^2=19600\) , \(150^2=22500\) , \(160^2=25600\) , \(170^2=28900 \) . So we see that \(28224\) is between \(160^2\) and \(170^2\) . Therefore, the number \(\sqrt(28224)\) is between \(160\) and \(170\) .
Let's try to determine the last digit. Let's remember what single-digit numbers when squaring give at the end \ (4 \) ? These are \(2^2\) and \(8^2\) . Therefore, \(\sqrt(28224)\) will end in either 2 or 8. Let's check this. Find \(162^2\) and \(168^2\) :
\(162^2=162\cdot 162=26224\)
\(168^2=168\cdot 168=28224\) .
Hence \(\sqrt(28224)=168\) . Voila!

In order to adequately solve the exam in mathematics, first of all, it is necessary to study the theoretical material, which introduces numerous theorems, formulas, algorithms, etc. At first glance, it may seem that this is quite simple. However, finding a source in which the theory for the Unified State Examination in mathematics is presented easily and understandably for students with any level of training is, in fact, a rather difficult task. School textbooks cannot always be kept at hand. And finding the basic formulas for the exam in mathematics can be difficult even on the Internet.

Why is it so important to study theory in mathematics, not only for those who take the exam?

1. Because it broadens your horizons. The study of theoretical material in mathematics is useful for anyone who wants to get answers to a wide range of questions related to the knowledge of the world. Everything in nature is ordered and has a clear logic. This is precisely what is reflected in science, through which it is possible to understand the world.
2. Because it develops the intellect. Studying reference materials for the exam in mathematics, as well as solving various problems, a person learns to think and reason logically, to formulate thoughts correctly and clearly. He develops the ability to analyze, generalize, draw conclusions.

We invite you to personally evaluate all the advantages of our approach to the systematization and presentation of educational materials.

The topic of square roots is mandatory in the school curriculum of the mathematics course. You can't do without them when solving quadratic equations. And later it becomes necessary not only to extract the roots, but also to perform other actions with them. Among them are quite complex: exponentiation, multiplication and division. But there are also quite simple ones: subtraction and addition of roots. By the way, they only seem so at first glance. Performing them without errors is not always easy for someone who is just starting to get acquainted with them.

What is a mathematical root?

This action arose as opposed to exponentiation. Mathematics assumes the presence of two opposite operations. There is subtraction for addition. Multiplication is opposed to division. The reverse action of the degree is the extraction of the corresponding root.

If the exponent is 2, then the root will be square. It is the most common in school mathematics. It does not even have an indication that it is square, that is, the number 2 is not assigned to it. The mathematical notation of this operator (radical) is shown in the figure.

From the described action, its definition follows smoothly. To extract the square root of a certain number, you need to find out what the radical expression will give when multiplied by itself. This number will be the square root. If we write this mathematically, we get the following: x * x \u003d x 2 \u003d y, which means √y \u003d x.

What actions can be taken with them?

At its core, a root is a fractional power that has a unit in the numerator. And the denominator can be anything. For example, the square root has a value of two. Therefore, all actions that can be performed with degrees will also be valid for roots.

And they have the same requirements for these actions. If multiplication, division and exponentiation do not meet with difficulties for students, then the addition of roots, as well as their subtraction, sometimes leads to confusion. And all because you want to perform these operations without looking at the sign of the root. And this is where the mistakes begin.

What are the rules for addition and subtraction?

First you need to remember two categorical "no":

• it is impossible to perform addition and subtraction of roots, as with prime numbers, that is, it is impossible to write the root expressions of the sum under one sign and perform mathematical operations with them;
• you cannot add and subtract roots with different exponents, such as square and cubic.

An illustrative example of the first ban: √6 + √10 ≠ √16 but √(6 + 10) = √16.

In the second case, it is better to limit ourselves to simplifying the roots themselves. And in the answer leave their sum.

Now to the rules

1. Find and group similar roots. That is, those who not only have the same numbers under the radical, but they themselves have one indicator.
2. Perform the addition of the roots combined into one group by the first action. It is easy to implement, because you only need to add the values ​​that come before the radicals.
3. Extract the roots in those terms in which the radical expression forms a whole square. In other words, do not leave anything under the sign of the radical.
4. Simplify root expressions. To do this, you need to factor them into prime factors and see if they give the square of any number. It is clear that this is true when it comes to the square root. When the exponent is three or four, then the prime factors must give the cube or the fourth power of the number.
5. Take out from under the sign of the radical a factor that gives an integer power.
6. See if similar terms appear again. If yes, then perform the second step again.

In a situation where the problem does not require the exact value of the root, it can be calculated on a calculator. Round off the infinite decimal fraction that will be displayed in its window. Most often this is done up to the hundredths. And then perform all operations for decimal fractions.

This is all the information about how the addition of the roots is performed. The examples below will illustrate the above.

First task

Calculate the value of expressions:

a) √2 + 3√32 + ½ √128 - 6√18;

b) √75 - √147 + √48 - 1/5 √300;

c) √275 - 10√11 + 2√99 + √396.

a) If you follow the algorithm above, you can see that there is nothing for the first two actions in this example. But you can simplify some radical expressions.

For example, factor 32 into two factors 2 and 16; 18 will be equal to the product of 9 and 2; 128 is 2 by 64. Given this, the expression will be written like this:

√2 + 3√(2 * 16) + ½ √(2 * 64) - 6 √(2 * 9).

Now you need to take out from under the radical sign those factors that give the square of the number. This is 16=4 2 , 9=3 2 , 64=8 2 . The expression will take the form:

√2 + 3 * 4√2 + ½ * 8 √2 - 6 * 3√2.

We need to simplify the writing a bit. For this, the coefficients are multiplied before the signs of the root:

√2 + 12√2 + 4 √2 - 12√2.

In this expression, all the terms turned out to be similar. Therefore, they just need to be folded. The answer will be: 5√2.

b) Like the previous example, the addition of roots begins with their simplification. The root expressions 75, 147, 48 and 300 will be represented by the following pairs: 5 and 25, 3 and 49, 3 and 16, 3 and 100. Each of them has a number that can be taken out from under the root sign:

5√5 - 7√3 + 4√3 - 1/5 * 10√3.

After simplification, the answer is: 5√5 - 5√3. It can be left in this form, but it is better to take the common factor 5 out of the bracket: 5 (√5 - √3).

c) And again factorization: 275 = 11 * 25, 99 = 11 * 9, 396 = 11 * 36. After factoring out the root sign, we have:

5√11 - 10√11 + 2 * 3√11 + 6√11. After reducing similar terms, we get the result: 7√11.

Fractional example

√(45/4) - √20 - 5√(1/18) - 1/6 √245 + √(49/2).

The following numbers will need to be factored: 45 = 5 * 9, 20 = 4 * 5, 18 = 2 * 9, 245 = 5 * 49. Similarly to those already considered, you need to take the factors out from under the root sign and simplify the expression:

3/2 √5 - 2√5 - 5/ 3 √(½) - 7/6 √5 + 7 √(½) = (3/2 - 2 - 7/6) √5 - (5/3 - 7 ) √(½) = - 5/3 √5 + 16/3 √(½).

This expression requires getting rid of the irrationality in the denominator. To do this, multiply the second term by √2/√2:

5/3 √5 + 16/3 √(½) * √2/√2 = - 5/3 √5 + 8/3 √2.

To complete the action, you need to select the integer part of the factors in front of the roots. The first is 1, the second is 2.

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The square root of the number x is the number a, which, when multiplied by itself, gives the number x: a * a = a^2 = x, ?x = a. As with any numbers, it is allowed to perform the arithmetic operations of addition and subtraction over square roots.

Instruction

1. First, when adding square roots, try to extract those roots. This will be valid if the numbers under the root sign are perfect squares. Let's say the expression?4 +?9 is given. The first number 4 is the square of the number 2. The second number 9 is the square of the number 3. So it turns out that: ?4 + ?9 = 2 + 3 = 5.

2. If there are no full squares under the root sign, then try to transfer the multiplier of the number from under the root sign. Let's say, let the expression? 24 +? 54 be given. Factorize the numbers: 24 = 2 * 2 * 2 * 3, 54 = 2 * 3 * 3 * 3. In the number 24 there is a factor 4, the one that can be transferred from the square root sign. The number 54 has a factor of 9. Thus, it turns out that: ?24 + ?54 = ?(4 * 6) +? (9 * 6) = 2 * ?6 + 3 * ?6 = 5 * ?6. In this example, as a result of removing the factor from the root sign, it turned out to simplify the given expression.

3. Let the sum of 2 square roots be the denominator of a fraction, say, A / (?a + ?b). And even if you are faced with the task of "getting rid of the irrationality in the denominator." Then you can use the next method. Multiply the numerator and denominator of the fraction by the expression ?a - ?b. Thus, in the denominator, you get the formula for abbreviated multiplication: (?a + ?b) * (?a - ?b) \u003d a - b. By analogy, if the difference of the roots is given in the denominator: ?a - ?b, then the numerator and denominator of the fraction must be multiplied by the expression?a + ?b. For example, let's say 4 / (?3 + ?5) = 4 * (?3 - ?5) / ((?3 + ?5) * (?3 - ?5)) = 4 * (?3 - ?5) / (-2) = 2 * (?5 - ?3).

4. Consider a more difficult example of getting rid of irrationality in the denominator. Let the fraction 12 / (?2 +?3 +?5) be given. You need to multiply the numerator and denominator of the fraction by the expression? 2 + ?3 - ?5:12 / (? 2 + ? + ?5) * (?2 + ?3 - ?5)) = 12 * (?2 + ?3 - ?5) / (2 * ?6) = ?6 * (?2 + ?3 - ?5) = 2 * ?3 + 3 * ?2 - ?30.

5. And finally, if you only need an approximate value, then you can calculate the square roots on the calculator. Calculate the values ​​separately for the whole number and write down with the required precision (say, two decimal places). And then perform the required arithmetic operations, as with ordinary numbers. Say, let's say you need to find out the approximate value of the expression? 7 +? 5 ? 2.65 + 2.24 = 4.89.

Related videos

Note!
In no case can square roots be added as primitive numbers, i.e. ?3 + ?2? ?5!!!

Useful advice
If you factor out the number in order to move the square from under the root sign, then do the reverse check - multiply all the resulting factors and get the original number.