The properties of the roots underlie the next two transformations, called bringing them under the root sign and taking them out from under the root sign, which we now consider.
Entering a multiplier under the sign of the root
Introducing a multiplier under the sign implies replacing the expression , where B and C are some numbers or expressions, and n is natural number, greater than one, is identically equal to an expression of the form or .
For example, an irrational expression after introducing a factor of 2 under the root sign takes the form .
Theoretical basis This transformation, the rules for its implementation, as well as solutions to various typical examples are given in the article introducing a multiplier under the sign of the root.
Removing the multiplier from under the root sign
A transformation, in a certain sense the opposite of introducing a factor under the root sign, is removing the factor from under the root sign. It consists of representing the root as a product for odd n or as a product for even n, where B and C are some numbers or expressions.
For an example, let’s return to the previous paragraph: the irrational expression, after removing the factor from under the root sign, takes the form . Another example: removing the factor from under the root sign in the expression gives the product, which can be rewritten as .
On what this transformation is based, and by what rules it is carried out, we will examine in a separate article the removal of the multiplier from under the sign of the root. There we will also give solutions to examples and list ways to reduce a radical expression to a form convenient for multiplying.
Converting fractions containing roots
Irrational expressions can contain fractions that have roots in the numerator and denominator. With such fractions you can carry out any of the basic identity transformations of fractions.
Firstly, nothing prevents you from working with expressions in the numerator and denominator. As an example, consider the fraction. The irrational expression in the numerator is obviously identically equal to , and by turning to the properties of roots, the expression in the denominator can be replaced by the root . As a result, the original fraction is converted to the form .
Second, you can change the sign in front of a fraction by changing the sign of the numerator or denominator. For example, the following transformations of an irrational expression take place: 
.
Thirdly, sometimes it is possible and advisable to reduce a fraction. For example, how to deny yourself the pleasure of reducing a fraction 
to the irrational expression, as a result we get 
.
It is clear that in many cases, before reducing a fraction, the expressions in its numerator and denominator have to be factored, which in simple cases allow you to achieve the abbreviated multiplication formula. And sometimes it helps to reduce a fraction by replacing a variable, which allows you to move from the original fraction with irrationality to a rational fraction, which is more comfortable and familiar to work with.
For example, let's take the expression . Let's introduce new variables and , in these variables the original expression has the form . Having performed in the numerator
Irrational expressions and their transformations
Last time we remembered (or learned, depending on who) what it is , learned how to extract such roots, sorted out the basic properties of roots piece by piece and solved simple examples with roots.
This lesson will be a continuation of the previous one and will be devoted to transformations of a wide variety of expressions containing all kinds of roots. Such expressions are called irrational. Expressions with letters, additional conditions, getting rid of irrationality in fractions, and some advanced techniques for working with roots will appear here. The techniques that will be discussed in this lesson, will be a good basis for solving Unified State Exam problems(and not only) of almost any level of complexity. So let's get started.
First of all, I will duplicate here the basic formulas and properties of roots. So as not to jump from topic to topic. Here they are:
at
You must know these formulas and be able to apply them. And in both directions - both from left to right and from right to left. It is on them that the solution to most tasks with roots of any degree of complexity is based. Let's start with the simplest thing for now - with the direct application of formulas or their combinations.
Easy application of formulas
In this part, simple and harmless examples will be considered - without letters, additional conditions and other tricks. However, even in them, as a rule, there are options. And the more sophisticated the example, the more such options there are. And an inexperienced student has the main problem– where to start? The answer here is simple - If you don't know what you need, do what you can. As long as your actions are in peace and harmony with the rules of mathematics and do not contradict them.) For example, this task:
Calculate:
Even in such a simple example, there are several possible paths to the answer.
The first is to simply multiply the roots by the first property and extract the root from the result:
The second option is this: we don’t touch it, we work with . We take the factor out from under the root sign, and then - according to the first property. Like this:
You can decide as much as you like. In any of the options, the answer is one - eight. For example, it’s easier for me to multiply 4 and 128 and get 512, and the cube root can be easily extracted from this number. If someone doesn’t remember that 512 is 8 cubed, then it doesn’t matter: you can write 512 as 2 9 (the first 10 powers of two, I hope you remember?) and using the formula for the root of the power:
Another example.
Calculate: .
If you work according to the first property (putting everything under one root), you will get a hefty number, from which the root can then be extracted - also not sugar. And it’s not a fact that it will be extracted exactly.) Therefore, it is useful here to remove the factors from under the root in the number. And make the most of:
And now everything is fine:
All that remains is to write the eight and two under one root (according to the first property) and the job is done. :)
Now let's add some fractions.
Calculate:
The example is quite primitive, but it also has options. You can use the multiplier to transform the numerator and reduce it with the denominator:
Or you can immediately use the formula for dividing roots:
As we see, this way and that – everything is correct.) If you don’t stumble halfway and make a mistake. Although where can I go wrong here...
Let us now look at the most recent example from homework last lesson:
Simplify:
A completely unimaginable set of roots, and even nested ones. What should I do? The main thing is not to be afraid! Here we first notice under the roots the numbers 2, 4 and 32 - powers of two. The first thing to do is to reduce all the numbers to twos: after all, the more identical numbers in the example and the fewer different ones, the easier it is.) Let’s start separately with the first factor:
The number can be simplified by reducing the two under the root with the four in the root exponent:
Now, according to the root of the work:
.
In the number we take out the two as the root sign:
And we deal with the expression using the root of the root formula:
So, the first factor will be written like this:
The nested roots have disappeared, the numbers have become smaller, which is already pleasing. It’s just that the roots are different, but we’ll leave it that way for now. If necessary, we will convert them to the same ones. Let's take the second factor.)
We transform the second factor in a similar way, using the formula of the root of the product and the root of the root. Where necessary, we reduce the indicators using the fifth formula:
We paste everything into the original example and get:
Got the product of a whole bunch of absolutely different roots. It would be nice to bring them all to one indicator, and then we’ll see. Well, it's quite possible. The largest of the root exponents is 12, and all the others - 2, 3, 4, 6 - are divisors of the number 12. Therefore, we will reduce all roots according to the fifth property to one exponent - 12:
We count and get:
We didn’t get a nice number, but that’s okay. We were asked simplify expression, not count. Simplified? Certainly! And the type of answer (integer or not) no longer plays any role here.
Some addition/subtraction and abbreviated multiplication formulas
Unfortunately, general formulas for adding and subtracting roots no in mathematics. However, in tasks these actions with roots are often found. Here it is necessary to understand that any roots are exactly the same mathematical symbols as letters in algebra.) And the same techniques and rules apply to roots as to letters - opening parentheses, bringing similar ones, abbreviated multiplication formulas, etc. P.
For example, it is clear to everyone that . Similar the same The roots can be added/subtracted to each other quite easily:
If the roots are different, then we look for a way to make them the same - by adding/subtracting a multiplier or using the fifth property. If it’s not simplified in any way, then perhaps the transformations are more cunning.
Let's look at the first example.
Find the meaning of the expression: .
All three roots, although cubic, are from different numbers. They are not purely extracted and are added/subtracted from each other. Therefore, the use of general formulas does not work here. What should I do? Let’s take out the factors in each root. In any case, it won’t be worse.) Moreover, there are, in fact, no other options:
That is, .
That's the solution. Here we moved from different roots to the same ones with the help removing the multiplier from under the root. And then they simply brought similar ones.) We decide further.
Find the value of an expression:
There's definitely nothing you can do about the root of seventeen. We work according to the first property - we make one root from the product of two roots:
Now let's take a closer look. What do we have under the big cube root? The difference is qua... Well, of course! Difference of squares:
Now all that remains is to extract the root: .
Calculate:
Here you will have to show mathematical ingenuity.) We think approximately as follows: “So, in the example, the product of roots. Under one root is the difference, and under the other is the sum. Very similar to the difference of squares formula. But... The roots are different! The first is square, and the second is of the fourth degree... It would be nice to make them the same. According to the fifth property, one can easily square root make the fourth root. To do this, it is enough to square the radical expression.”
If you thought about the same, then you are halfway to success. Absolutely right! Let's turn the first factor into a fourth root. Like this:
Now, there is nothing to be done, but you will have to remember the formula for the square of the difference. Only when applied to roots. So what? Why are roots worse than other numbers or expressions?! We build:
“Hmm, well, they erected it, so what? Horseradish is not sweeter than radish. Stop! And if you take out the four under the root? Then the same expression will emerge as under the second root, only with a minus, and this is exactly what we are trying to achieve!”
Right! Let's take four:
.
And now - a matter of technology:
This is how complex examples are untangled.) Now it's time to practice with fractions.
Calculate:
It is clear that the numerator must be converted. How? Using the formula of the square of the sum, of course. Do we have any other options? :) We square it, take out the factors, reduce the indicators (where necessary):
Wow! We got exactly the denominator of our fraction.) This means that the whole fraction is obviously equal to one:
Another example. Only now on another formula for abbreviated multiplication.)
Calculate:
It is clear that the square of the difference must be used in practice. We write out the denominator separately and - let's go!
We take out the factors from under the roots:
Hence,
Now everything bad is superbly reduced and it turns out:
Well, let's take it to the next level. :)
Letters and additional conditions
Literal expressions with roots are a trickier thing than numeric expressions, and is an inexhaustible source of annoying and very serious mistakes. Let's close this source.) Errors arise due to the fact that such tasks often involve negative numbers and expressions. They are either given to us directly in the task, or hidden in letters and additional conditions. And in the process of working with roots, we constantly need to remember that in the roots even degree both under the root itself and as a result of root extraction there should be non-negative expression. The key formula in the tasks of this paragraph will be the fourth formula:
There are no questions with roots of odd degrees - everything is always extracted, both positive and negative. And the minus, if anything, is brought forward. Let's get straight to the roots even degrees.) For example, such a short task.
Simplify: , If .
It would seem that everything is simple. It will just turn out to be X.) But why then the additional condition? In such cases, it is useful to estimate with numbers. Purely for myself.) If, then x is obviously a negative number. Minus three, for example. Or minus forty. Let . Can you raise minus three to the fourth power? Certainly! The result is 81. Is it possible to extract the fourth root of 81? Why not? Can! You get three. Now let's analyze our entire chain:
What do we see? The input was a negative number, and the output was already positive. It was minus three, now it’s plus three.) Let’s return to the letters. Without a doubt, modulo it will be exactly X, but only X itself is minus (by condition!), and the result of extraction (due to the arithmetic root!) must be plus. How to get a plus? Very simple! To do this, it is enough to put a minus in front of a obviously negative number.) And correct solution looks like that:
By the way, if we used the formula, then, remembering the definition of a module, we would immediately get the correct answer. Because the
|x| = -x at x<0.
Take the factor out of the root sign: , Where .
The first glance is at the radical expression. Everything is OK here. In any case, it will be non-negative. Let's start extracting. Using the formula for the root of a product, we extract the root of each factor:
I don’t think there’s any need to explain where the modules came from.) Now let’s analyze each of the modules.
Multiplier | a | we leave it unchanged: we don’t have any conditions for the lettera. We don't know whether it's positive or negative. Next module |b 2 | can be safely omitted: in any case, the expressionb 2 non-negative. But about |c 3 | - there’s already a problem here.) If, then c 3 <0. Стало быть, модуль надо раскрыть with a minus: | c 3 | = - c 3 . In total, the correct solution would be:
And now - the reverse problem. Not the easiest, I warn you right away!
Enter a multiplier under the sign of the root: .
If you immediately write down the solution like this
then you fell into a trap. This wrong decision! What's the matter?
Let's take a closer look at the expression under the root. Under the root of the fourth power, as we know, there should be non-negative expression. Otherwise, the root has no meaning.) Therefore And this, in turn, means that and, therefore, itself is also non-positive: .
And the mistake here is that we are introducing at the root non-positive number: the fourth degree turns it into non-negative and the wrong result is obtained - on the left there is a deliberate minus, and on the right there is already a plus. And put it at the root even degree we have the right only non-negative numbers or expressions. And leave the minus, if there is one, in front of the root.) How can we identify a non-negative factor in the number, knowing that it itself is completely negative? Yes, exactly the same! Put a minus.) And so that nothing changes, compensate for it with another minus. Like this:
And now already non-negative We calmly enter the number (-b) under the root according to all the rules:
This example clearly shows that, unlike other branches of mathematics, in the roots the correct answer does not always follow automatically from the formulas. You need to think and personally make the right decision.) You should especially be more careful with the signs in irrational equations and inequalities.
Let's look at the next important technique when working with roots - getting rid of irrationality.
Eliminating irrationality in fractions
If the expression contains roots, then, let me remind you, such an expression is called expression with irrationality. In some cases, it can be useful to get rid of this very irrationality (i.e. roots). How can you eliminate the root? Our root disappears when... raised to a power. With an indicator either equal to the root indicator or a multiple of it. But, if we raise the root to a power (i.e. multiply the root by itself the required number of times), then the expression will change. Not good.) However, in mathematics there are topics where multiplication is quite painless. In fractions, for example. According to the basic property of a fraction, if the numerator and denominator are multiplied (divided) by the same number, the value of the fraction will not change.
Let's say we are given this fraction:
Is it possible to get rid of the root in the denominator? Can! To do this, the root must be cubed. What are we missing in the denominator for a full cube? We are missing a multiplier, i.e.. So we multiply the numerator and denominator of the fraction by
The root in the denominator has disappeared. But... he appeared in the numerator. Nothing can be done, such is fate.) This is no longer important to us: we were asked to free the denominator from the roots. Released? Undoubtedly.)
By the way, those who are already comfortable with trigonometry may have paid attention to the fact that in some textbooks and tables, for example, they designate differently: somewhere , and somewhere . The question is - what is right? Answer: everything is correct!) If you guess that– this is simply the result of liberation from irrationality in the denominator of the fraction. :)
Why should we free ourselves from irrationality in fractions? What difference does it make whether the root is in the numerator or in the denominator? The calculator will calculate everything anyway.) Well, for those who do not part with a calculator, there is really practically no difference... But even counting on a calculator, you can pay attention to the fact that divide on whole number is always more convenient and faster than on irrational. And I’ll keep silent about division into a column.)
The following example will only confirm my words.
How can we eliminate the square root of the denominator here? If the numerator and denominator are multiplied by the expression, then the denominator will be the square of the sum. The sum of the squares of the first and second numbers will give us just numbers without any roots, which is very pleasing. However... it will pop up double product the first number to the second, where the root of three will still remain. It doesn't channel. What should I do? Remember another wonderful formula for abbreviated multiplication! Where there are no double products, but only squares:
An expression that, when multiplied by a certain sum (or difference), produces difference of squares, also called conjugate expression. In our example, the conjugate expression will be the difference. So we multiply the numerator and denominator by this difference:
What can I say? As a result of our manipulations, not only did the root of the denominator disappear, but the fraction disappeared altogether! :) Even with a calculator, subtracting the root of three from a three is easier than calculating a fraction with the root in the denominator. Another example.
Free yourself from irrationality in the denominator of a fraction:
How to get out of this? Formulas for abbreviated multiplication with squares do not work right away - it will not be possible to completely eliminate the roots due to the fact that this time our root is not square, but cubic. It is necessary that the root is somehow raised into a cube. Therefore, one of the formulas with cubes must be used. Which one? Let's think about it. The denominator is the sum. How can we achieve the cube of the root? Multiply by partial squared difference! So, we will apply the formula sum of cubes. This one:
As a we have three, and as a quality b– cube root of five:
And again the fraction disappeared.) Such situations, when, when freed from irrationality in the denominator of a fraction, the fraction itself completely disappears along with the roots, occur very often. How do you like this example!
Calculate:
Just try adding these three fractions! No mistakes! :) One common denominator is worth it. What if you try to free yourself from the irrationality in the denominator of each fraction? Well, let's try:
Wow, how interesting! All the fractions are gone! Completely. And now the example can be solved in two ways:
Simple and elegant. And without long and tedious calculations. :)
That is why one must be able to do the operation of liberation from irrationality in fractions. In such sophisticated examples, it’s the only thing that saves, yes.) Of course, no one canceled attentiveness. There are tasks where you are asked to get rid of irrationality in numerator. These tasks are no different from those discussed, only the numerator is cleared from the roots.)
More complex examples
It remains to consider some special techniques for working with roots and practice unraveling not the simplest examples. And then the information received will be enough to solve tasks with roots of any level of complexity. So - go ahead.) First, let's figure out what to do with nested roots when the root from root formula does not work. For example, here's an example.
Calculate:
The root is under the root... Moreover, under the roots is the sum or difference. Therefore, the formula for the root of the root (with multiplication of exponents) is here It does not work. So something needs to be done about radical expressions: We simply have no other options. In such examples, most often the large root is encrypted perfect square some amount. Or differences. And the root of the square is already perfectly extracted! And now our task is to decrypt it.) Such decryption is beautifully done through system of equations. Now you will see everything for yourself.)
So, under the first root we have this expression:
What if you didn’t guess right? Let's check! We square it using the formula for the square of the sum:
That's right.) But... Where did I get this expression from? From the sky?
No.) We will get it a little lower honestly. Simply using this expression, I show exactly how task writers encrypt such squares. :) What is 54? This sum of squares of the first and second numbers. And, pay attention, already without roots! And the root remains in double product, which in our case is equal to . Therefore, unraveling such examples begins with searching for the double product. If you unravel with the usual selection. And, by the way, about signs. Everything is simple here. If there is a plus before the double, then the square of the sum. If it’s a minus, then the differences.) We have a plus – which means the square of the sum.) And now – the promised analytical method of decoding. Through the system.)
So, under our root there is clearly hanging out the expression (a+b) 2, and our task is to find a And b. In our case, the sum of squares gives 54. So we write:
Now double the product. We have it. So we write it down:
We got this system:
We solve by the usual substitution method. We express from the second equation, for example, and substitute it into the first:
Let's solve the first equation:
Got biquadratic equation relativea . We calculate the discriminant:
Means,
We got as many as four possible valuesa. We are not afraid. Now we will weed out all the unnecessary things.) If we now calculate the corresponding values for each of the four found values, we will get four solutions to our system. Here they are:
And here the question is - which solution is right for us? Let's think about it. Negative solutions can be immediately discarded: when squaring, the minuses will “burn out”, and the entire radical expression as a whole will not change.) The first two options remain. You can choose them completely arbitrarily: rearranging the terms still does not change the sum.) Let, for example, , a .
In total, we got the square of the following sum under the root:
Everything is clear.)
It’s not for nothing that I describe the decision process in such detail. To make it clear how decryption occurs.) But there is one problem. The analytical method of decoding, although reliable, is very long and cumbersome: you have to solve a biquadratic equation, get four solutions to the system and then still think about which ones to choose... Troubling? I agree, it's troublesome. This method works flawlessly in most of these examples. However, very often you can save yourself a lot of work and find both numbers creatively. By selection.) Yes, yes! Now, using the example of the second term (second root), I will show an easier and faster way to isolate the complete square under the root.
So now we have this root: 
.
Let's think like this: “Under the root is most likely an encrypted complete square. Once there is a minus before the double, it means the square of the difference. The sum of the squares of the first and second numbers gives us the number 54. But what kind of squares are these? 1 and 53? 49 and 5 ? There are too many options... No, it’s better to start untangling with double the product. Ourcan be written as . Times product doubled, then we immediately discard the two. Then candidates for the role a and b remain 7 and . What if it's 14 and/2 ? It's possible. But we always start with something simple!” So, let , a . Let's check them for the sum of squares:
Happened! This means that our radical expression is actually the square of the difference:
Here is a light way to avoid messing with the system. It doesn't always work, but in many of these examples it is quite sufficient. So, under the roots there are complete squares. All that remains is to correctly extract the roots and calculate the example:
Now let’s look at an even more non-standard task on roots.)
Prove that the number A– integer, if .
Nothing is directly extracted, the roots are embedded, and even of different degrees... A nightmare! However, the task makes sense.) Therefore, there is a key to solving it.) And the key here is this. Consider our equality
How equation relative A. Yes Yes! It would be nice to get rid of the roots. Our roots are cubic, so let’s cube both sides of the equation. According to the formula cube of the sum:
Cubes and cubic roots cancel each other out, and under each large root we take one bracket from the square and collapse the product of the difference and the sum into a difference of squares:
Separately, we calculate the difference of squares under the roots:
Expressions containing a radical sign (root) are called irrational.
An arithmetic root of a natural power $n$ of a non-negative number a is some non-negative number that, when raised to the power $n$, produces the number $a$.
$(√^n(a))^n=a$
In the notation $√^n(a)$, “a” is called the radical number, $n$ is the exponent of the root or radical.
Properties of $n$th roots for $a≥0$ and $b≥0$:
1. The root of the product is equal to the product of the roots
$√^n(a∙b)=√^n(a)∙√^n(b)$
Calculate $√^5(5)∙√^5(625)$
The root of a product is equal to the product of roots and vice versa: the product of roots with the same root exponent is equal to the root of the product of radical expressions
$√^n(a)∙√^n(b)=√^n(a∙b)$
$√^5{5}∙√^5{625}=√^5{5∙625}=√^5{5∙5^4}=√^5{5^5}=5$
2. The root of a fraction is a separate root from the numerator and a separate root from the denominator
$√^n((a)/(b))=(√^n(a))/(√^n(b))$, for $b≠0$
3. When a root is raised to a power, the radical expression is raised to this power
$(√^n(a))^k=√^n(a^k)$
4. If $a≥0$ and $n,k$ are natural numbers greater than $1$, then the equality is true.
$√^n(√^k(a))=√^(n∙k)a$
5. If the indicators of the root and radical expression are multiplied or divided by the same natural number, then the value of the root will not change.
$√^(n∙m)a^(k∙m)=√^n(a^k)$
6. The root of an odd degree can be extracted from positive and negative numbers, and the root of an even degree - only from positive ones.
7. Any root can be represented as a power with a fractional (rational) exponent.
$√^n(a^k)=a^((k)/(n))$
Find the value of the expression $(√(9∙√^11(s)))/(√^11(2048∙√s))$ for $s>0$
The root of the product is equal to the product of the roots
$(√(9∙√^11(s)))/(√^11(2048∙√s))=(√9∙√(√^11(s)))/(√^11(2048)∙ √^11(√с))$
We can extract roots from numbers immediately
$(√9∙√(√^11(s)))/(√^11(2048)∙√^11(√s))=(3∙√(√^11(s)))/(2∙ √^11(√с))$
$√^n(√^k(a))=√^(n∙k)a$
$(3∙√(√^11(s)))/(2∙√^11(√s))=(3∙√^22(s))/(2∙√^22(s))$
We reduce the $22$ roots of $с$ and get $(3)/(2)=1.5$
Answer: $1.5$
If for a radical with an even exponent we do not know the sign of the radical expression, then when extracting the root, the module of the radical expression comes out.
Find the value of the expression $√((с-7)^2)+√((с-9)^2)$ at $7< c < 9$
If there is no exponent above the root, this means that we are working with a square root. Its indicator is two, i.e. honest. If for a radical with an even exponent we do not know the sign of the radical expression, then when extracting the root, the module of the radical expression comes out.
$√((с-7)^2)+√((с-9)^2)=|c-7|+|c-9|$
Let's determine the sign of the expression under the modulus sign based on the condition $7< c < 9$
To check, take any number from a given range, for example, $8$
Let's check the sign of each module
$8-9<0$, при раскрытии модуля пользуемся правилом: модуль положительного числа равен самому себе, отрицательного числа - равен противоположному значению. Так как у второго модуля знак отрицательный, при раскрытии меняем знак перед модулем на противоположный.
$|c-7|+|c-9|=(с-7)-(с-9)=с-7-с+9=2$
Properties of powers with rational exponent:
1. When multiplying powers with the same bases, the base remains the same, and the exponents are added.
$a^n∙a^m=a^(n+m)$
2. When raising a degree to a power, the base remains the same, but the exponents are multiplied
$(a^n)^m=a^(n∙m)$
3. When raising a product to a power, each factor is raised to this power
$(a∙b)^n=a^n∙b^n$
4. When raising a fraction to a power, the numerator and denominator are raised to this power
The article reveals the meaning of irrational expressions and transformations with them. Let's consider the very concept of irrational expressions, transformation and characteristic expressions.
Yandex.RTB R-A-339285-1
What are irrational expressions?
When introducing roots at school, we study the concept of irrational expressions. Such expressions are closely related to roots.
Definition 1
Irrational expressions are expressions that have a root. That is, these are expressions that have radicals.
Based on this definition, we have that x - 1, 8 3 3 6 - 1 2 3, 7 - 4 3 (2 + 3) , 4 a 2 d 5: d 9 2 a 3 5 - these are all expressions of an irrational type.
When considering the expression x · x - 7 · x + 7 x + 3 2 · x - 8 3 we find that the expression is rational. Rational expressions include polynomials and algebraic fractions. Irrational ones include working with logarithmic expressions or radical expressions.
Main types of transformations of irrational expressions
When calculating such expressions, it is necessary to pay attention to the DZ. Often they require additional transformations in the form of opening parentheses, bringing similar members, groupings, and so on. The basis of such transformations is operations with numbers. Transformations of irrational expressions adhere to a strict order.
Example 1
Transform the expression 9 + 3 3 - 2 + 4 · 3 3 + 1 - 2 · 3 3 .
Solution
It is necessary to replace the number 9 with an expression containing the root. Then we get that
81 + 3 3 - 2 + 4 3 3 + 1 - 2 3 3 = = 9 + 3 3 - 2 + 4 3 3 + 1 - 2 3 3
The resulting expression has similar terms, so let's perform the reduction and grouping. We get
9 + 3 3 - 2 + 4 3 3 + 1 - 2 3 3 = = 9 - 2 + 1 + 3 3 + 4 3 3 - 2 3 3 = = 8 + 3 3 3
Answer: 9 + 3 3 - 2 + 4 3 3 + 1 - 2 3 3 = 8 + 3 3 3
Example 2
Present the expression x + 3 5 2 - 2 · x + 3 5 + 1 - 9 as a product of two irrationals using abbreviated multiplication formulas.
Solutions
x + 3 5 2 - 2 x + 3 5 + 1 - 9 = = x + 3 5 - 1 2 - 9
We represent 9 in the form of 3 2, and we apply the formula for the difference of squares:
x + 3 5 - 1 2 - 9 = x + 3 5 - 1 2 - 3 2 = = x + 3 5 - 1 - 3 x + 3 5 - 1 + 3 = = x + 3 5 - 4 x + 3 5 + 2
The result of identical transformations led to the product of two rational expressions that needed to be found.
Answer:
x + 3 5 2 - 2 x + 3 5 + 1 - 9 = = x + 3 5 - 4 x + 3 5 + 2
You can perform a number of other transformations that apply to irrational expressions.
Converting a Radical Expression
The important thing is that the expression under the root sign can be replaced by one that is identically equal to it. This statement makes it possible to work with a radical expression. For example, 1 + 6 can be replaced by 7 or 2 · a 5 4 - 6 by 2 · a 4 · a 4 - 6 . They are identically equal, so the replacement makes sense.
When there is no a 1 different from a, where an inequality of the form a n = a 1 n is valid, then such an equality is possible only for a = a 1. The values of such expressions are equal to any values of the variables.
Using Root Properties
The properties of roots are used to simplify expressions. To apply the property a · b = a · b, where a ≥ 0, b ≥ 0, then from the irrational form 1 + 3 · 12 can become identically equal to 1 + 3 · 12. Property. . . a n k n 2 n 1 = a n 1 · n 2 · , . . . , · n k , where a ≥ 0 means that x 2 + 4 4 3 can be written in the form x 2 + 4 24 .
There are some nuances when converting radical expressions. If there is an expression, then - 7 - 81 4 = - 7 4 - 81 4 we cannot write it down, since the formula a b n = a n b n serves only for non-negative a and positive b. If the property is applied correctly, then the result will be an expression of the form 7 4 81 4 .
For correct transformation, transformations of irrational expressions using the properties of roots are used.
Entering a multiplier under the sign of the root
Definition 3Place under the root sign- means to replace the expression B · C n, and B and C are some numbers or expressions, where n is a natural number that is greater than 1, with an equal expression that looks like B n · C n or - B n · C n.
If we simplify the expression of the form 2 x 3, then after adding it to the root, we get that 2 3 x 3. Such transformations are possible only after a detailed study of the rules for introducing a multiplier under the root sign.
Removing the multiplier from under the root sign
If there is an expression of the form B n · C n , then it is reduced to the form B · C n , where there are odd n , which take the form B · C n with even n , B and C being some numbers and expressions.
That is, if we take an irrational expression of the form 2 3 x 3, remove the factor from under the root, then we get the expression 2 x 3. Or x + 1 2 · 7 will result in an expression of the form x + 1 · 7, which has another notation of the form x + 1 · 7.
Removing the multiplier from under the root is necessary to simplify the expression and quickly convert it.
Converting fractions containing roots
An irrational expression can be either a natural number or a fraction. To convert fractional expressions, pay great attention to its denominator. If we take a fraction of the form (2 + 3) x 4 x 2 + 5 3, then the numerator will take the form 5 x 4, and, using the properties of the roots, we find that the denominator will become x 2 + 5 6. The original fraction can be written as 5 x 4 x 2 + 5 6.
It is necessary to pay attention to the fact that it is necessary to change the sign of only the numerator or only the denominator. We get that
X + 2 x - 3 x 2 + 7 4 = x + 2 x - (- 3 x 2 + 7 4) = x + 2 x 3 x 2 - 7 4
Reducing a fraction is most often used when simplifying. We get that
3 · x + 4 3 - 1 · x x + 4 3 - 1 3 reduce by x + 4 3 - 1 . We get the expression 3 x x + 4 3 - 1 2.
Before reduction, it is necessary to perform transformations that simplify the expression and make it possible to factorize a complex expression. Abbreviated multiplication formulas are most often used.
If we take a fraction of the form 2 · x - y x + y, then it is necessary to introduce new variables u = x and v = x, then the given expression will change form and become 2 · u 2 - v 2 u + v. The numerator should be expanded into polynomials according to the formula, then we get that
2 · u 2 - v 2 u + v = 2 · (u - v) · u + v u + v = 2 · u - v . After performing the reverse substitution, we arrive at the form 2 x - y, which is equal to the original one.
Reduction to a new denominator is allowed, then it is necessary to multiply the numerator by an additional factor. If we take a fraction of the form x 3 - 1 0, 5 · x, then we reduce it to the denominator x. to do this, you need to multiply the numerator and denominator by the expression 2 x, then we get the expression x 3 - 1 0, 5 x = 2 x x x 3 - 1 0, 5 x 2 x = 2 x x 3 - 1 x .
Reducing fractions or bringing similar ones is necessary only for the ODZ of the specified fraction. When we multiply the numerator and denominator by an irrational expression, we find that we get rid of the irrationality in the denominator.
Getting rid of irrationality in the denominator
When an expression gets rid of the root in the denominator by transformation, it is called getting rid of irrationality. Let's look at the example of a fraction of the form x 3 3. After getting rid of irrationality, we obtain a new fraction of the form 9 3 x 3.
Transition from roots to powers
Transitions from roots to powers are necessary for quickly transforming irrational expressions. If we consider the equality a m n = a m n , we can see that its use is possible when a is a positive number, m is an integer, and n is a natural number. If we consider the expression 5 - 2 3, then otherwise we have the right to write it as 5 - 2 3. These expressions are equivalent.
When the root contains a negative number or a number with variables, then the formula a m n = a m n is not always applicable. If you need to replace such roots (- 8) 3 5 and (- 16) 2 4 with powers, then we get that - 8 3 5 and - 16 2 4 by the formula a m n = a m n we do not work with negative a. In order to analyze in detail the topic of radical expressions and their simplifications, it is necessary to study the article on the transition from roots to powers and back. It should be remembered that the formula a m n = a m n is not applicable for all expressions of this type. Getting rid of irrationality contributes to further simplification of the expression, its transformation and solution.
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