A general look at the conversion of fractions. Competent transformation of rational expressions Transformation of fractionally rational expressions

Rational expressions and fractions are the cornerstone of the entire course of algebra. Those who learn how to work with such expressions, simplify them and factor them, in fact, will be able to solve any problem, since the transformation of expressions is an integral part of any serious equation, inequality, and even a word problem.

In this video tutorial, we'll see how to correctly apply abbreviated multiplication formulas to simplify rational expressions and fractions. Let's learn to see these formulas where, at first glance, there is nothing. At the same time, we repeat such a simple trick as factoring a square trinomial into factors through the discriminant.

As you probably already guessed from the formulas behind my back, today we will study the formulas for abbreviated multiplication, or rather, not the formulas themselves, but their application to simplify and reduce complex rational expressions. But, before moving on to solving examples, let's take a closer look at these formulas or recall them:

1. $((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right)$ is the difference of squares;
2. $((\left(a+b \right))^(2))=((a)^(2))+2ab+((b)^(2))$ is the square of the sum;
3. $((\left(a-b \right))^(2))=((a)^(2))-2ab+((b)^(2))$ is the squared difference;
4. $((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b)^( 2)) \right)$ is the sum of cubes;
5. $((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^(2) ) \right)$ is the difference of cubes.

I would also like to note that our school education system is designed in such a way that it is with the study of this topic, i.e. rational expressions, as well as roots, modules, all students have the same problem, which I will now explain.

The fact is that at the very beginning of studying the formulas for abbreviated multiplication and, accordingly, actions to reduce fractions (this is about grade 8), teachers say something like this: “If something is not clear to you, then don’t worry, we will we will return to this topic more than once, in high school for sure. We'll figure it out later." Well, then at the turn of grades 9-10, the same teachers explain to the same students who still don’t know how to solve rational fractions, something like this: “Where were you the previous two years? The same was studied in algebra in the 8th grade! What can be incomprehensible here? It's so obvious!"

However, for ordinary students, such explanations do not make it any easier: they both had a mess in their head, and they still have it, so right now we will analyze two simple examples, on the basis of which we will see how to single out these expressions in real problems, which will lead us to abbreviated multiplication formulas and how to then apply this to transform complex rational expressions.

Reduction of simple rational fractions

Task #1

\[\frac(4x+3((y)^(2)))(9((y)^(4))-16((x)^(2)))\]

The first thing we need to learn is to distinguish exact squares and higher powers in the original expressions, on the basis of which we can then apply the formulas. Let's get a look:

Let's rewrite our expression taking into account these facts:

\[\frac(4x+3((y)^(2)))(((\left(3((y)^(2)) \right))^(2))-((\left(4x \right))^(2)))=\frac(4x+3((y)^(2)))(\left(3((y)^(2))-4x \right)\left(3 ((y)^(2))+4x \right))=\frac(1)(3((y)^(2))-4x)\]

Answer: $\frac(1)(3((y)^(2))-4x)$.

Task #2

Let's move on to the second task:

\[\frac(8)(((x)^(2))+5xy-6((y)^(2)))\]

There is nothing to simplify here, because the numerator is a constant, but I proposed this problem precisely so that you learn how to factorize polynomials containing two variables. If instead of it there was a polynomial written below, how would we decompose it?

\[((x)^(2))+5x-6=\left(x-... \right)\left(x-... \right)\]

Let's solve the equation and find $x$ that we can put in place of dots:

\[((x)^(2))+5x-6=0\]

\[((x)_(1))=\frac(-5+7)(2)=\frac(2)(2)=1\]

\[((x)_(2))=\frac(-5-7)(2)=\frac(-12)(2)=-6\]

We can rewrite the trinomial as follows:

\[((x)^(2))+5xy-6((y)^(2))=\left(x-1 \right)\left(x+6 \right)\]

We learned how to work with a square trinomial - for this we had to record this video lesson. But what if, in addition to $x$ and the constant, there is also $y$? Let's look at them as another element of the coefficients, i.e. Let's rewrite our expression as follows:

\[((x)^(2))+5y\cdot x-6((y)^(2))\]

\[((x)_(1))=\frac(-5y+7y)(2)=y\]

\[((x)_(2))=\frac(-5y-7y)(2)=\frac(-12y)(2)=-6y\]

We write the decomposition of our square construction:

\[\left(x-y \right)\left(x+6y \right)\]

In total, if we return to the original expression and rewrite it taking into account the changes, we get the following:

\[\frac(8)(\left(x-y \right)\left(x+6y \right))\]

What does such a record give us? Nothing, because it cannot be reduced, it is not multiplied or divided by anything. However, as soon as this fraction turns out to be an integral part of a more complex expression, such an expansion will come in handy. Therefore, as soon as you see a square trinomial (whether it is burdened with additional parameters or not), always try to factor it.

Nuances of the solution

Remember the basic rules for converting rational expressions:

• All denominators and numerators must be factored either through abbreviated multiplication formulas or through the discriminant.
• We need to work according to this algorithm: when we look and try to highlight the abbreviated multiplication formula, then, first of all, we try to translate everything to the maximum possible degree. After that, we take the general degree out of brackets.
• Very often there will be expressions with a parameter: other variables will appear as coefficients. We find them using the quadratic expansion formula.

Thus, as soon as you see rational fractions, the first thing to do is to factor both the numerator and denominator into factors (into linear expressions), while we use the reduced multiplication formulas or the discriminant.

Let's look at a couple of such rational expressions and try to factor them out.

Solving More Complex Examples

Task #1

\[\frac(4((x)^(2))-6xy+9((y)^(2)))(2x-3y)\cdot \frac(9((y)^(2))- 4((x)^(2)))(8((x)^(3))+27((y)^(3)))\]

We rewrite and try to expand each term:

Let's rewrite our entire rational expression with these facts in mind:

\[\frac(((\left(2x \right))^(2))-2x\cdot 3y+((\left(3y \right))^(2)))(2x-3y)\cdot \frac (((\left(3y \right))^(2))-((\left(2x \right))^(2)))(((\left(2x \right))^(3))+ ((\left(3y\right))^(3)))=\]

\[=\frac(((\left(2x \right))^(2))-2x\cdot 3y+((\left(3y \right))^(2)))(2x-3y)\cdot \ frac(\left(3y-2x \right)\left(3y+2x \right))(\left(2x+3y \right)\left(((\left(2x \right))^(2))- 2x\cdot 3y+((\left(3y \right))^(2)) \right))=-1\]

Answer: $-1$.

Task #2

\[\frac(3-6x)(2((x)^(2))+4x+8)\cdot \frac(2x+1)(((x)^(2))+4-4x)\ cdot \frac(8-((x)^(3)))(4((x)^(2))-1)\]

Let's look at all fractions.

\[((x)^(2))+4-4x=((x)^(2))-4x+2=((x)^(2))-2\cdot 2x+((2)^( 2))=((\left(x-2 \right))^(2))\]

Let's rewrite the whole structure taking into account the changes:

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

\[=\frac(3\cdot \left(-1 \right))(2\cdot \left(x-2 \right)\cdot \left(-1 \right))=\frac(3)(2 \left(x-2 \right))\]

Answer: $\frac(3)(2\left(x-2 \right))$.

Nuances of the solution

So what have we just learned:

• Not every square trinomial is factorized, in particular, this applies to the incomplete square of the sum or difference, which are very often found as parts of the sum or difference cubes.
• Constants, i.e. ordinary numbers that do not have variables with them can also act as active elements in the decomposition process. Firstly, they can be taken out of brackets, and secondly, the constants themselves can be represented as powers.
• Very often, after decomposing all elements into factors, opposite constructions arise. You need to reduce these fractions very carefully, because when you cross them out either from above or from below, an additional factor $-1$ appears - this is precisely the consequence of the fact that they are opposite.

Solving complex problems

\[\frac(27((a)^(3))-64((b)^(3)))(((b)^(2))-4):\frac(9((a)^ (2))+12ab+16((b)^(2)))(((b)^(2))+4b+4)\]

Let's consider each term separately.

First fraction:

\[((\left(3a \right))^(3))-((\left(4b \right))^(3))=\left(3a-4b \right)\left(((\left (3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2)) \right)\]

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

We can rewrite the entire numerator of the second fraction as follows:

\[((\left(3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2))\]

Now let's look at the denominator:

\[((b)^(2))+4b+4=((b)^(2))+2\cdot 2b+((2)^(2))=((\left(b+2 \right ))^(2))\]

Let's rewrite the entire rational expression with the above facts in mind:

\[\frac(\left(3a-4b \right)\left(((\left(3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2 )) \right))(\left(b-2 \right)\left(b+2 \right))\cdot \frac(((\left(b+2 \right))^(2)))( ((\left(3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2)))=\]

\[=\frac(\left(3a-4b \right)\left(b+2 \right))(\left(b-2 \right))\]

Answer: $\frac(\left(3a-4b \right)\left(b+2 \right))(\left(b-2 \right))$.

Nuances of the solution

As we have seen once again, the incomplete squares of the sum or the incomplete squares of the difference, which are often found in real rational expressions, however, do not be afraid of them, because after the transformation of each element they almost always cancel. In addition, in no case should you be afraid of large constructions in the final answer - it is quite possible that this is not your mistake (especially if everything is factored), but the author conceived such an answer.

In conclusion, I would like to analyze one more complex example, which is no longer directly related to rational fractions, but it contains everything that awaits you on real tests and exams, namely: factorization, reduction to common denominator, cancellation of like terms. That's exactly what we're going to do now.

Solving a complex problem of simplifying and transforming rational expressions

\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]

First, consider and expand the first bracket: in it we see three separate fractions with different denominators, so the first thing we need to do is bring all three fractions to a common denominator, and for this, each of them should be factored:

\[((x)^(2))+2x+4=((x)^(2))+2\cdot x+((2)^(2))\]

\[((x)^(2))-8=((x)^(3))-((2)^(2))=\left(x-2 \right)\left(((x) ^(2))+2x+((2)^(2)) \right)\]

Let's rewrite our entire structure as follows:

\[\frac(x)(((x)^(2))+2x+((2)^(2)))+\frac(((x)^(2))+8)(\left(x -2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))-\frac(1)(x-2)=\]

\[=\frac(x\left(x-2 \right)+((x)^(3))+8-\left(((x)^(2))+2x+((2)^(2 )) \right))(\left(x-2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))=\]

\[=\frac(((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))=\frac(((x)^(2))-4x-4)(\ left(x-2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))=\]

\[=\frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+(( 2)^(2)) \right))=\frac(x-2)(((x)^(2))+2x+4)\]

This is the result of the calculations from the first parenthesis.

Dealing with the second parenthesis:

\[((x)^(2))-4=((x)^(2))-((2)^(2))=\left(x-2 \right)\left(x+2 \ right)\]

Let's rewrite the second bracket, taking into account the changes:

\[\frac(((x)^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)=\frac( ((x)^(2))+2\left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^ (2))+2x+4)(\left(x-2 \right)\left(x+2 \right))\]

Now let's write the entire original construction:

\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]

Answer: $\frac(1)(x+2)$.

Nuances of the solution

As you can see, the answer turned out to be quite sane. However, please note: very often with such large-scale calculations, when the only variable is only in the denominator, students forget that this is the denominator and it should be at the bottom of the fraction and write this expression in the numerator - this is a gross mistake.

In addition, I would like to draw your special attention to how such tasks are formalized. In any complex calculations, all steps are performed step by step: first, we count the first bracket separately, then the second bracket separately, and only at the end do we combine all the parts and calculate the result. Thus, we insure ourselves against stupid mistakes, carefully write down all the calculations and at the same time do not waste any extra time, as it might seem at first glance.

Teaching without coercion

(A guide to the fascinating world of mathematics)

Mathematics already then needs to be taught, that it puts the mind in order. (M.V. Lomonosov)

So how do you learn math?

This question interests many.

The first step is to close the gaps from the past. If you missed (didn’t understand, didn’t study in principle, etc.) any topic, sooner or later you will definitely step on this rake. With a classic result... That's the way mathematics works.

Whether you're learning a new topic or revisiting an old one, master the math definitions and terms! Pay attention, I do not say - "learn", but I say "master". These are different things. You must understand, for example, what is the denominator, discriminant, or arcsine at a simple, even primitive level. What is it, why is it needed and how to deal with it. Life will become easier.

If I ask you how to use the dense restricted environment transition device, you will feel uncomfortable answering, right? And if you understand that this very device is an ordinary door? It's actually kind of more fun.

And, of course, you need to decide. If you don't know how to decide, no big deal. You have to try and try. All once did not know how. But those who tried and tried, albeit incorrectly, with mistakes, now know how to solve. And who did not try, did not study - he never learned.

Here are the three components of the answer to the question: "How to teach mathematics?" Eliminate gaps, master the terms at an understandable level and meaningfully solve tasks.

If mathematics seems to you a jungle of some rules, formulas, expressions in which it is impossible to navigate, then I will console you. There are paths and guiding stars there! You will settle in, get used to it, and you will also begin to admire these wilds ...

The mathematics of the school course does not solve complex examples because he can't. She can well solve something like 5x \u003d 10, a quadratic equation through the discriminant, and the same simple one from trigonometry, logarithms, etc. And all the power of mathematics is aimed at simplifying complex expressions. It is for this that rules and formulas for various transformations are needed. They allow us to write the original expression in a different form convenient for us without changing its essence.

"Mathematics is the art of calling different things by the same name." (A. Poincare)

For example, 8 = 6 + 2 = 2 = = log 6561 = 32: 4. It's still the same number 8! Only recorded in a variety of forms. Which type to choose - we decide! Consistent with the task and common sense.

The main guiding star in mathematics is the ability to transform expressions. Almost any solution starts with a transformation of the original expression. With the help of rules and formulas, which are not at all such an insane amount as you think.

We often say "All formulas work from left to right and from right to left." Let's say (a + b) almost everyone writes it down as a + 2ab + b . But not everyone (unfortunately) realizes that x + 2x + 1 can be written as (x + 1) . And here's what you need to know! Formulas need to know in person! To be able to recognize them in expressions encrypted by cunning teachers, to identify parts of the formulas, to bring, if necessary, to complete ones.

Expression conversions are troublesome at first. Requires labor. At the initial stage, it is necessary to check, where possible, the correctness of the transformation by inverse transformation. Factored out - multiply back and bring similar ones. It turned out the original expression - hurray! Found the roots of the equation - substitute in the original expression. See what happened. And so on.

So I invite you to wonderful world mathematics. And let's start our journey by getting to know fractions, as this is perhaps the most vulnerable spot for most schoolchildren.

Good luck!

Lesson 1.

Types of fractions. Transformations.

Who knows fractions, he is strong, he is brave in mathematics!

Fractions are of three types.

1. Common fractions , For example: , , , .

Sometimes, instead of a horizontal line, they put a slash: 1/2, 3/7, 19/5. A line, both horizontal (vinculium) and oblique (solidus) means the same operation: dividing the top number (numerator) by the bottom number (denominator). And that's it! Instead of a line, it is quite possible to put a division sign - two dots. 1/2 = 1:2.

When the division is possible entirely, it must be done. So, instead of the fraction 32/8, it is much more pleasant to write the number 4. That is. 32 is simply divided by 8. 32/8 = 32: 8 = 4. I'm not talking about the fraction 4/1, which is also equal to 4. And if it doesn’t divide completely, we leave it as a fraction. Sometimes you have to do the reverse. Make a fraction from a whole number. But more on that later.

2. Decimals , for example: 0.5; 3.28; 0.543; 23.32.

3. mixed numbers , For example: , , , .

Mixed numbers are practically not used in high school. In order to work with them, they must be converted to ordinary fractions. But you definitely need to know how to do it! And then such a number will come across in the task and hang ... From scratch. But we remember this procedure!

Common fractions are the most versatile. Let's start with them. By the way, if there are all sorts of logarithms, sines and other letters in the fraction, this does not change anything. In the sense that all actions with fractional expressions are no different from actions with ordinary fractions!

So go ahead! The whole variety of fraction transformations is provided by a single property! That's what it's called basic property of a fraction. Remember: if the numerator and denominator of a fraction are multiplied (divided) by the same number, the fraction will not change. Those:

And we need it, all these transformations? - you ask. And how! Now you will see for yourself. First, let's use the basic property of a fraction to reduce fractions. It would seem that the thing is elementary. We divide the numerator and denominator by the same number and that's it! It's impossible to go wrong! But... man is a creative being. You can make mistakes everywhere! Especially if you have to reduce not a fraction of the form 5/10, but a fractional rational expression.

Usually the student does not think about dividing the numerator and denominator by the same number (or expression)! He just crosses out everything the same from above and below! This is where it hides typical mistake, blooper if you want.

For example, you need to simplify the expression: .

What are we doing? We cross out the factor a above and the degree below! We get: .

Everything is correct. But really you shared the whole numerator And the whole denominator on multiplier a. If you are used to just crossing out, then, in a hurry, you can cross out the letter a in the expression and get again. Which would be categorically wrong: an unforgivable mistake. Because here the whole numerator on a already not shared! This fraction cannot be reduced.

When reducing, you need to divide the entire numerator and the entire denominator!

Reducing fractions makes life a lot easier. You will get a fraction somewhere, for example, 375/1000. And how to work with her now? Without a calculator? Multiply, say, add, square!? And if you are not too lazy, but carefully reduce by five, and even by five, and even ... while it is being reduced. We get 3/8! Much nicer, right?

The main property of a fraction allows you to convert ordinary fractions to decimals and vice versa, without a calculator! It's important in CT, right?

It's easy with decimals. As it is heard, so it is written! Let's say 0.25. It's zero point, twenty-five hundredths. So we write: 25/100. We reduce (divide the numerator and denominator by 25), we get an ordinary fraction: 1/4. All. It happens, and nothing is reduced. For example, 0.3. This is three tenths, i.e. 3/10.

What if integers are non-zero? It's OK. We write the whole fraction without any commas in the numerator, and in the denominator - what is heard. For example: 3.17. This is three whole, seventeen hundredths. We write 317 in the numerator, and 100 in the denominator. We get 317/100. Nothing is reduced, that means everything. This is the answer. From all of the above, a useful conclusion: Any decimal fraction can be converted to a common fraction.

But the reverse conversion, ordinary to decimal, some cannot do without a calculator. And it is necessary! How are you going to write down the answer? We carefully read and master this process.

What is a decimal fraction? Her denominator is always 10, or 100, or 1000, or 10,000, and so on. If your common fraction has such a denominator, there is no problem. For example, 4/10 = 0.4. Or 7/100 = 0.07. Or 12/10 = 1.2. What if the result is 1/2? And the answer must be written in decimal ...

We remember basic property of a fraction! Mathematics favorably allows you to multiply the numerator and denominator by the same number. For anyone, by the way! Except zero, of course. Let's use this feature to our advantage! What can the denominator be multiplied by, i.e. 2 so that it becomes 10, or 100, or 1000 (smaller is better, of course...)? 5, obviously. Feel free to multiply the denominator by 5. But then the numerator must also be multiplied by 5. We get 1/2 = 0.5. That's all.

However, the denominators may be different. For example, the fraction 3/16. Then you can simply divide 3 by 16. In the absence of a calculator, you will have to divide by a corner, as they taught in elementary grades. We get 0.1875.

And there are some very bad denominators. For example, the fraction 1/3 cannot be turned into a good decimal. And on the calculator, and when dividing by a corner, we get 0.3333333 ... Hence, one more useful conclusion. Not every common fraction converts to a decimal!

So, with ordinary and decimal fractions sorted out. It remains to deal with mixed numbers. To work with them, they need to be converted to ordinary fractions. How to do it? You can catch a fifth grader and ask him. But not always a fifth-grader will be nearby ... You will have to do it yourself. It is not difficult. Multiply the denominator of the fractional part by the integer part and add the numerator of the fractional part. This will be the numerator of a common fraction. What about the denominator? The denominator will remain the same. It sounds complicated, but it's actually quite simple. Let's see an example.

Suppose that in the task you saw a number with horror:

Calmly, without panic, we argue. The whole part is 1. One. The fractional part is 3/7. Therefore, the denominator of the fractional part is 7. This denominator will be the denominator of the ordinary fraction. Consider: numerator. We multiply 7 by 1 (the integer part) and add 3 (the numerator of the fractional part). We get 10. This will be the numerator of an ordinary fraction. That's all. It looks even simpler in mathematical notation:

Easily? Then secure your success! Convert these mixed numbers , , to common fractions. You should get 10/3, 23/10 and 21/4.

Well, almost everything. You remembered the types of fractions and understood how to translate them from one type to another. The question remains: why do this? Where and when to apply this deep knowledge?

Any example itself suggests the necessary actions. If in the example ordinary fractions, decimals, and even mixed numbers are mixed into a bunch, we translate everything into ordinary fractions. It can always be done. Well, if it is written, for example, 0.8 + 0.3, then we think so, without any translation. Why do we need extra work? We choose the way to solve which is convenient for us!

If the task is full of decimal fractions, but um ... some scary ones, go to ordinary ones, try it! Maybe everything will work out. For example, you have to square the number 0.125. Not so easy if you have not lost the habit of the calculator! Not only do you need to multiply the numbers in a column, but also think about where to insert the comma! It certainly doesn't work in my mind! And if you go to an ordinary fraction? 0.125 = 125/1000. We reduce by 5 (this is for starters). We get 25/200. Once again on 5. We get 5/40. Still shrinking! Back to 5! We get 1/8. Easily square (in your mind!) and get 1/64. All!

Let's summarize our lesson.

1. There are three types of fractions: ordinary, decimal and mixed numbers.

2. Decimals and mixed numbers can always be converted to common fractions. Reverse transfer is not always possible.

3. The choice of the type of fractions for working with the task depends on this very task. If there are different types of fractions in one task, the most reliable thing is to switch to ordinary fractions.

Practical Tips:

1. The most important thing when working with fractional expressions is accuracy and attentiveness! Is not common words, not good wishes! This is a severe need! It is better to write two extra lines in a draft than to make a mistake when calculating in your head.

2. In the examples with different types fractions - go to ordinary fractions.

3. We reduce all fractions to the stop.

4. We reduce multi-level fractional expressions to ordinary ones using division through two points (we follow the order of division!).

5. We divide the unit into a fraction in our mind, simply by turning the fraction over.

Now try to put the theory into practice.

So, let's solve it in the exam mode! We solve an example, we check, we solve the following. We decided everything - we checked again from the first to the last example. And then we look at the answers.

Decided? Looking for answers that match yours. The answers are written in disorder, away from temptation, so to speak...

0; 17/22; 3; 1; 3/4; 14; -5/4; 17/12; 1/3; 5; 2/5; 25.

And now we draw conclusions. If everything worked out - happy for you! Elementary calculations with fractions are not your problem! You can do more serious things. If not... Patience and work will grind everything.

This generalized material is known from the school mathematics course. Here we consider fractions of a general form with numbers, powers, roots, logarithms, trigonometric functions, or other objects. The basic transformations of fractions will be considered, regardless of their type.

What is a fraction?

Definition 1

There are several more definitions.

Definition 2

The horizontal slash that separates A and B is called the fraction or fractional line.

Definition 3

The expression above the bar of a fraction is called numerator and under - denominator.

From ordinary fractions to general fractions

Acquaintance with a fraction occurs in the 5th grade, when ordinary fractions pass. It can be seen from the definition that the numerator and denominator are natural numbers.

Example 1

For example 1 5 , 2 6 , 12 7 , 3 1 , which can be written as 1 / 5 , 2 / 6 , 12 / 7 , 3 / 1 .

After studying operations with ordinary fractions, we deal with fractions that have not one natural number in the denominator, but expressions with natural numbers.

Example 2

For example, 1 + 3 5 , 9 - 5 16 , 2 7 9 12 .

When we are dealing with fractions, where there are letters or literal expressions, it is written as follows:

a + b c , a - b c , a c b d .

Definition 4

Fix the rules for addition, subtraction, multiplication of ordinary fractions a c + b c = a + b c , a c - b c = a - b c , a b v d = a c b d

To calculate, it is often necessary to come to the translation of mixed numbers into ordinary fractions. When we denote the integer part as a, then the fractional part has the form b / c, we get a fraction of the form a · c + b c, from which it is clear the appearance of such fractions 2 · 11 + 3 11 , 5 · 2 + 1 2 and so on.

The line of a fraction is regarded as a sign of division. Therefore, the record can be converted in another way:

1: a - (2 b + 1) = 1 a - 2 b + 1, 5 - 1, 7 3: 2 3 - 4: 2 = 5 - 1, 7 3 2 3 - 4: 2 , where the quotient 4: 2 can be replaced by a fraction, then we get an expression of the form

5 - 1 , 7 3 2 3 - 4 2

Calculations with rational fractions occupy a special place in mathematics, since the numerator and denominator can contain not just numerical values, but polynomials.

Example 3

For example, 1 x 2 + 1 , x y - 2 y 2 0 , 5 - 2 x + y 3 .

Rational expressions are considered as fractions of a general form.

Example 4

For example, x x + 1 4 x 2 x 2 - 1 2 x 3 + 3 , 1 + x 2 y (x - 2) 1 x + 3 x 1 + 2 - x 4 x 5 + 6x.

The study of roots, powers with rational exponents, logarithms, trigonometric functions suggests that their application appears in given fractions of the form:

Example 5

a n b n , 2 x + x 2 3 x 1 3 - 12 x , 2 x 2 + 3 3 x 2 + 3 , ln (x - 3) ln e 5 , cos 2 α - sin 2 α 1 - 1 cos 2 α .

Fractions can be combined, that is, have the form x + 1 x 3 log 3 sin 2 x + 3, lg x + 2 lg x 2 - 2 x + 1.

Types of fraction conversions

For a number of identical transformations, several types are considered:

Definition 5

• transformation specific to working with the numerator and denominator;
• sign change before a fractional expression;
• reduction to a common denominator and fraction reduction;
• representation of a fraction as a sum of polynomials.

Converting Expressions in the Numerator and Denominator

Definition 6

With identically equal expressions, we have that the resulting fraction is identically equal to the original.

If a fraction of the form A / B is given, then A and B are some expressions. Then, when replacing, we get a fraction of the form A 1 / B 1 . It is necessary to prove the equality A / A 1 = B / B 1 for any value of variables that satisfy the ODZ.

We have that A And A 1 And B And B1 are identically equal, then their values ​​are also equal. It follows that for any value A/B And A 1 / B 1 fractions will be equal.

This conversion makes it easier to work with fractions if you need to convert the numerator and the denominator separately.

Example 6

For example, let's take a fraction of the form 2 / 18, which we convert to 2 2 · 3 · 3. To do this, we decompose the denominator into prime factors. The fraction x 2 + x y x 2 + 2 x y + y 2 \u003d x x + y (x + y) 2 has a numerator of the form x 2 + x y, means that it is necessary to replace with x (x + y) , which will be obtained by bracketing the common factor x . The denominator of a given fraction x 2 + 2 x y + y 2 collapse by the abbreviated multiplication formula. Then we get that its identically equal expression is (x + y) 2 .

Example 7

If a fraction of the form sin 2 3 φ - π + cos 2 3 φ - π φ φ 5 6 is given, then to simplify it is necessary to replace the numerator by 1 according to the formula, and bring the denominator to the form φ 11 12. Then we get that 1 φ 11 12 is equal to the given fraction.

Change of sign in front of a fraction, in its numerator, denominator

Fraction conversions are also the replacement of signs in front of the fraction. Let's look at some rules:

Definition 7

• when changing the sign of the numerator, we get a fraction that is equal to the given one, and it literally looks like _ - A - B \u003d A B, where A and B are some expressions;
• when changing the sign before the fraction and before the numerator, we get that - - A B = A B ;
• when replacing the sign in front of the fraction and its denominator, we get that - A - B = A B .

Proof

The minus sign is in most cases treated as a signed factor - 1 , and the slash is division. From here we get that - A - B = - 1 · A: - 1 · B . Grouping the factors, we have that

1 A: - 1 B = ((- 1) : (- 1) A: B = = 1 A: B = A: B = A B

After proving the first assertion, we justify the rest. We get:

A B = (- 1) (((- 1) A) : B) = (- 1 - 1) A: B = = 1 (A: B) = A: B = A B - A - B = (- 1) (A: - 1 B) = ((- 1) : (- 1)) (A: B) == 1 (A: B) = A: B = A B

Consider examples.

Example 8

When it is necessary to convert the fraction 3/7 to the form - 3 - 7, - - 3 7, - 3 - 7, then it is similarly performed with a fraction of the form - 1 + x - x 2 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x .

The transformations are performed as follows:

1) - 1 + x - x 2 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x = = - (- 1 + x - x 2) - 2 2 3 - ln x 2 + 3 x + sin 2 x 3 x = = 1 - x + x 2 - 2 2 3 + ln (x 2 + 3) x - s i n 2 x 3 x 2) - 1 + x - x 2 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x = = - - (- 1 + x - x 2) 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x = = - 1 - x + x 2 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x 3) - 1 + x - x 2 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x = = - - 1 + x - x 2 - 2 2 3 - ln (x 2 + 3) x + sin 2 x 3 x = = - - 1 + x - x 2 - 2 2 3 + ln (x 2 + 3) x - sin 2 x 3 x

Bringing a fraction to a new denominator

When studying ordinary fractions, we touched on the basic property of fractions, which allows you to multiply, divide the numerator and denominator by the same natural number. This can be seen from the equality a · m b · m = a b and a: m b: m = a b , where a , b , m are natural numbers.

This equality is valid for any values ​​a , b , m and all a except b ≠ 0 and m ≠ 0 . That is, we get that if the numerator of the fraction A / B with A and C, which are some expressions, is multiplied or divided by the expression M, not equal to 0, then we get a fraction that is identically equal to the initial one. We get that A · M B · M = A B and A: M B: M = A B .

This shows that the transformations are based on 2 transformations: reduction to a common denominator, reduction.

When reducing to a common denominator, multiplication is performed by the same number or expression, numerator and denominator. That is, we move on to solving the identical equal converted fraction.

Consider examples.

Example 9

If we take the fraction x + 1 0, 5 x 3 and multiply by 2, then we get that the new denominator will be 2 x 0, 5 x 3 = x 3, and the expression will take the form 2 x + 1 x 3.

Example 10

To reduce the fraction 1 - x 2 x 2 3 1 + ln x to another denominator of the form 6 x 1 + ln x 3, the numerator and denominator must be multiplied by 3 x 1 3 (1 + ln x) 2. As a result, we get the fraction 3 x 1 3 1 + ln x 2 1 - x 6 x (1 + ln x) 3

Such a transformation as getting rid of the irrationality in the denominator is also applicable. It eliminates the presence of a root in the denominator, which simplifies the solution process.

Fraction reduction

The main property is a transformation, that is, its direct reduction. When reduced, we get a simplified fraction. Let's look at an example:

Example 11

Or a fraction of the form x 3 x 3 x 2 (2 x 2 + 1 + 3) x 3 x 3 2 x 2 + 1 + 3 3 + 1 3 x, where the reduction is made using x 3, x 3 , 2 x 2 + 1 + 3 or an expression like x 3 x 3 2 x 2 + 1 + 3 . Then we get the fraction x 2 3 + 1 3 x

Fraction reduction is simple when the common factors are immediately visible. In practice, this does not occur often, therefore, it is first necessary to carry out some transformations of expressions of this kind. There are cases when it is necessary to find a common factor.

If there is a fraction of the form x 2 2 3 (1 - cos 2 x) 2 sin x 2 cos x 2 2 x 1 3, then it is necessary to apply trigonometric formulas and the properties of powers in order to be able to convert the fraction to the form x 1 3 x 2 1 3 sin 2 x sin 2 x x 1 3 . This will make it possible to reduce it by x 1 3 · sin 2 x .

Representing a fraction as a sum

When the numerator has an algebraic sum of expressions like A 1 , A 2 , … , A n, and the denominator is denoted B, then this fraction can be represented as A 1 / B , A 2 / B , … , A n / B.

Definition 8

To do this, fix this A 1 + A 2 + . . . + A n B = A 1 B + A 2 B + . . . + A n B .

This transformation is fundamentally different from adding fractions with the same exponents. Consider an example.

Example 12

Given a fraction of the form sin x - 3 x + 1 + 1 x 2, which we will represent as an algebraic sum of fractions. To do this, imagine as sin x x 2 - 3 x + 1 x 2 + 1 x 2 or sin x - 3 x + 1 x 2 + 1 x 2 or sin x x 2 + - 3 x + 1 + 1 x 2.

Any fraction that has the form A / B is represented as a sum of fractions in any way. The expression A in the numerator can be reduced or increased by any number or expression A 0 that will make it possible to get to A + A 0 B - A 0 B .

The decomposition of a fraction into the simplest is a special case for converting a fraction into a sum. Most often it is used in complex calculations for integration.

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The material of this article is a general look at the transformation of expressions containing fractions. Here we will consider the basic transformations that are characteristic of expressions with fractions.

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Fractional expressions and fractional expressions

To begin with, let's clarify what kind of expression transformation we are going to deal with.

The title of the article contains the self-explanatory phrase " expressions with fractions". That is, below we will talk about the transformation of numeric expressions and expressions with variables, in the record of which there is at least one fraction.

We note right away that after the publication of the article " Transformation of fractions: a general view"We are no longer interested in individual fractions. Thus, further we will consider sums, differences, products, partial and more complex expressions with roots, powers, logarithms, which are united only by the presence of at least one fraction.

And let's talk about fractional expressions. This is not the same as expressions with fractions. Fraction expressions are a more general concept. Not every expression with fractions is a fractional expression. For example, the expression is not a fractional expression, although it contains a fraction, it is an integer rational expression. So don't call an expression with fractions a fractional expression without being completely sure that it is.

Basic identical transformations of expressions with fractions

Example.

Simplify the expression .

Solution.

In this case, you can open the brackets, which will give the expression , which contains like terms and , as well as −3 and 3 . After their reduction, we get a fraction.

Let's show a short form of writing the solution:

Answer:

.

Working with individual fractions

The expressions we are talking about transforming differ from other expressions mainly in the presence of fractions. And the presence of fractions requires tools to work with them. In this paragraph, we will discuss the transformation of individual fractions included in the record of this expression, and in the next paragraph we will proceed to perform operations with the fractions that make up the original expression.

With any fraction that is a component of the original expression, you can perform any of the transformations indicated in the article Fraction conversion. That is, you can take a separate fraction, work with its numerator and denominator, reduce it, bring it to a new denominator, etc. It is clear that with this transformation, the selected fraction will be replaced by a fraction identically equal to it, and the original expression will be replaced by an expression identically equal to it. Let's look at an example.

Example.

Convert expression with fraction to a simpler form.

Solution.

Let's start the transformation by working with a fraction. First, open the brackets and give similar terms in the numerator of the fraction: . Now it begs the bracketing of the common factor x in the numerator and the subsequent reduction of the algebraic fraction: . It remains only to substitute the result obtained instead of a fraction in the original expression, which gives .

Answer:

.

Performing actions with fractions

Part of the process of converting expressions with fractions is often to do actions with fractions. They are carried out in accordance with the accepted procedure for performing actions. It is also worth keeping in mind that any number or expression can always be represented as a fraction with a denominator of 1.

Example.

Simplify the expression .

Solution.

The problem can be approached from different angles. In the context of the topic under consideration, we will go by performing actions with fractions. Let's start by multiplying fractions:

Now we write the product as a fraction with a denominator 1, after which we subtract the fractions:

If desired and necessary, one can still get rid of irrationality in the denominator , on which you can finish the transformation.

Answer:

Application of properties of roots, powers, logarithms, etc.

The class of expressions with fractions is very wide. Such expressions, in addition to the actual fractions, may contain roots, degrees with different exponents, modules, logarithms, trigonometric functions and so on. Naturally, when they are converted, the corresponding properties are applied.

Applicable to fractions, it is worth highlighting the property of the root of the fraction, the property of the fraction to the degree, the property of the modulus of the quotient and the property of the logarithm of the difference .

For clarity, we give a few examples. For example, in the expression It may be useful, based on the properties of the degree, to replace the first fraction with a degree, which further allows us to represent the expression as a squared difference. When converting a logarithmic expression it is possible to replace the logarithm of a fraction with the difference of logarithms, which further allows us to bring similar terms and thereby simplify the expression: . Converting trigonometric expressions may require replacing the ratio of the sine to the cosine of the same angle with a tangent. You may also have to move from the half argument using the appropriate formulas to the whole argument, thereby getting rid of the fraction argument, for example, .

Applying properties of roots, degrees, etc. to the transformation of expressions is covered in more detail in the articles:

• Transformation of irrational expressions using properties of roots,
• Transformation of expressions using the properties of powers,
• Converting logarithmic expressions using the properties of logarithms,
• Converting trigonometric expressions.

Decimal numbers such as 0.2; 1.05; 3.017 etc. as they are heard, so they are written. Zero point two, we get a fraction. One whole five hundredths, we get a fraction. Three whole seventeen thousandths, we get a fraction. The digits before the decimal point in a decimal number are the integer part of the fraction. The number after the decimal point is the numerator of the future fraction. If there is a one-digit number after the decimal point, the denominator will be 10, if two-digit - 100, three-digit - 1000, etc. Some of the resulting fractions can be reduced. In our examples

Converting a fraction to a decimal number

This is the reverse of the previous transformation. What is a decimal fraction? Her denominator is always 10, or 100, or 1000, or 10,000, and so on. If your usual fraction has such a denominator, there is no problem. For example, or

If a fraction, for example . In this case, you need to use the basic property of the fraction and convert the denominator to 10 or 100, or 1000 ... In our example, if we multiply the numerator and denominator by 4, we get a fraction that can be written as a decimal number 0.12.

Some fractions are easier to divide than to convert the denominator. For example,

Some fractions cannot be converted to decimal numbers!
For example,

Converting a mixed fraction to an improper

A mixed fraction, such as , can easily be converted to an improper fraction. To do this, you need to multiply the integer part by the denominator (bottom) and add it to the numerator (top), leaving the denominator (bottom) unchanged. That is

When converting a mixed fraction to an improper one, you can remember that you can use the addition of fractions

Converting an improper fraction to a mixed one (highlighting the whole part)

An improper fraction can be converted to a mixed fraction by highlighting the whole part. Consider an example, . Determine how many integer times "3" fit in "23". Or we divide 23 by 3 on the calculator, the whole number up to the decimal point is the desired one. This is "7". Next, we determine the numerator of the future fraction: we multiply the resulting "7" by the denominator "3" and subtract the result from the numerator "23". How would we find the excess that remains from the numerator "23", if we remove the maximum number of "3". The denominator is left unchanged. Everything is done, write down the result