Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school:

log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) − g (x)) (k (x) − 1) ∨ 0

Instead of the “∨” checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same.

This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the range of acceptable values. If you have forgotten the ODZ of a logarithm, I strongly recommend repeating it - see “What is a logarithm”.

Everything related to the range of acceptable values ​​must be written out and solved separately:

f(x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1.

These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values ​​has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready.

Task. Solve the inequality:

First, let’s write out the logarithm’s ODZ:

The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have:

x 2 + 1 ≠ 1;
x2 ≠ 0;
x ≠ 0.

It turns out that the ODZ of the logarithm is all numbers except zero: x ∈ (−∞ 0)∪(0; +∞). Now we solve the main inequality:

We make the transition from logarithmic inequality to rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” sign. We have:

(10 − (x 2 + 1)) · (x 2 + 1 − 1)< 0;
(9 − x 2) x 2< 0;
(3 − x) · (3 + x) · x 2< 0.

The zeros of this expression are: x = 3; x = −3; x = 0. Moreover, x = 0 is a root of the second multiplicity, which means that when passing through it, the sign of the function does not change. We have:

We get x ∈ (−∞ −3)∪(3; +∞). This set is completely contained in the ODZ of the logarithm, which means this is the answer.

Converting logarithmic inequalities

Often the original inequality is different from the one above. This can be easily corrected using the standard rules for working with logarithms - see “Basic properties of logarithms”. Namely:

1. Any number can be represented as a logarithm with a given base;
2. The sum and difference of logarithms with the same bases can be replaced by one logarithm.

Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows:

1. Find the VA of each logarithm included in the inequality;
2. Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms;
3. Solve the resulting inequality using the scheme given above.

Task. Solve the inequality:

Let's find the domain of definition (DO) of the first logarithm:

We solve using the interval method. Finding the zeros of the numerator:

3x − 2 = 0;
x = 2/3.

Then - the zeros of the denominator:

x − 1 = 0;
x = 1.

We mark zeros and signs on the coordinate arrow:

We get x ∈ (−∞ 2/3)∪(1; +∞). The second logarithm will have the same VA. If you don't believe it, you can check it. Now we transform the second logarithm so that the base is two:

As you can see, the threes at the base and in front of the logarithm have been reduced. We got two logarithms with the same base. Let's add them up:

log 2 (x − 1) 2< 2;
log 2 (x − 1) 2< log 2 2 2 .

We obtained the standard logarithmic inequality. We get rid of logarithms using the formula. Since the original inequality contains a “less than” sign, the resulting rational expression must also be less than zero. We have:

(f (x) − g (x)) (k (x) − 1)< 0;
((x − 1) 2 − 2 2)(2 − 1)< 0;
x 2 − 2x + 1 − 4< 0;
x 2 − 2x − 3< 0;
(x − 3)(x + 1)< 0;
x ∈ (−1; 3).

We got two sets:

1. ODZ: x ∈ (−∞ 2/3)∪(1; +∞);
2. Candidate answer: x ∈ (−1; 3).

It remains to intersect these sets - we get the real answer:

We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get x ∈ (−1; 2/3)∪(1; 3) - all points are punctured.

For example, the inequality is the expression \(x>5\).

Types of inequalities:

If \(a\) and \(b\) are numbers or , then the inequality is called numerical. It's actually just comparing two numbers. Such inequalities are divided into faithful And unfaithful.

For example:
\(-5<2\) - верное числовое неравенство, ведь \(-5\) действительно меньше \(2\);

\(17+3\geq 115\) is an incorrect numerical inequality, since \(17+3=20\), and \(20\) is less than \(115\) (and not greater than or equal to).

If \(a\) and \(b\) are expressions containing a variable, then we have inequality with variable. Such inequalities are divided into types depending on the content:

\(2x+1\geq4(5-x)\)				Variable only to the first power
\(3x^2-x+5>0\)				There is a variable in the second power (square), but there are no higher powers (third, fourth, etc.)
\(\log_(4)((x+1))<3\)
\(2^(x)\leq8^(5x-2)\)

... and so on.

What is the solution to an inequality?

If you substitute a number instead of a variable into an inequality, it will turn into a numeric one.

If a given value for x turns the original inequality into a true numerical one, then it is called solution to inequality. If not, then this value is not a solution. And to solve inequality– you need to find all its solutions (or show that there are none).

For example, if we substitute the number \(7\) into the linear inequality \(x+6>10\), we get the correct numerical inequality: \(13>10\). And if we substitute \(2\), there will be an incorrect numerical inequality \(8>10\). That is, \(7\) is a solution to the original inequality, but \(2\) is not.

However, the inequality \(x+6>10\) has other solutions. Indeed, we will get the correct numerical inequalities when substituting \(5\), and \(12\), and \(138\)... And how can we find all possible solutions? For this they use For our case we have:

\(x+6>10\) \(|-6\)
\(x>4\)

That is, any number greater than four is suitable for us. Now you need to write down the answer. Solutions to inequalities are usually written numerically, additionally marking them on the number axis with shading. For our case we have:

Answer: \(x\in(4;+\infty)\)

When does the sign of an inequality change?

There is one big trap in inequalities that students really “love” to fall into:

When multiplying (or dividing) an inequality by a negative number, it is reversed (“more” by “less”, “more or equal” by “less than or equal”, and so on)

Why is this happening? To understand this, let's look at the transformations of the numerical inequality \(3>1\). It is correct, three is indeed greater than one. First, let's try to multiply it by any positive number, for example, two:

\(3>1\) \(|\cdot2\)
\(6>2\)

As we can see, after multiplication the inequality remains true. And no matter what positive number we multiply by, we will always get the correct inequality. Now let’s try to multiply by a negative number, for example, minus three:

\(3>1\) \(|\cdot(-3)\)
\(-9>-3\)

The result is an incorrect inequality, because minus nine is less than minus three! That is, in order for the inequality to become true (and therefore, the transformation of multiplication by negative was “legal”), you need to reverse the comparison sign, like this: \(−9<− 3\).
With division it will work out the same way, you can check it yourself.

The rule written above applies to all types of inequalities, not just numerical ones.

Example: Solve the inequality \(2(x+1)-1<7+8x\)
Solution:

\(2x+2-1<7+8x\)	Let's move \(8x\) to the left, and \(2\) and \(-1\) to the right, not forgetting to change the signs
\(2x-8x<7-2+1\)
\(-6x<6\) \(|:(-6)\)	Let's divide both sides of the inequality by \(-6\), not forgetting to change from “less” to “more”
	Let's mark a numerical interval on the axis. Inequality, therefore we “prick out” the value \(-1\) itself and do not take it as an answer
	Let's write the answer as an interval

Answer: \(x\in(-1;\infty)\)

Inequalities and disability

Inequalities, just like equations, can have restrictions on , that is, on the values ​​of x. Accordingly, those values ​​that are unacceptable according to the DZ should be excluded from the range of solutions.

Example: Solve the inequality \(\sqrt(x+1)<3\)

Solution: It is clear that in order for the left side to be less than \(3\), the radical expression must be less than \(9\) (after all, from \(9\) just \(3\)). We get:

\(x+1<9\) \(|-1\)
\(x<8\)

All? Any value of x smaller than \(8\) will suit us? No! Because if we take, for example, the value \(-5\) that seems to fit the requirement, it will not be a solution to the original inequality, since it will lead us to calculating the root of a negative number.

\(\sqrt(-5+1)<3\)
\(\sqrt(-4)<3\)

Therefore, we must also take into account the restrictions on the value of X - it cannot be such that there is a negative number under the root. Thus, we have the second requirement for x:

\(x+1\geq0\)
\(x\geq-1\)

And for x to be the final solution, it must satisfy both requirements at once: it must be less than \(8\) (to be a solution) and greater than \(-1\) (to be admissible in principle). Plotting it on the number line, we have the final answer:

Answer: \(\left[-1;8\right)\)

Objective of the lesson: consider solving more complex inequalities.

During the classes

I. Statement of the topic and purpose of the lesson.

II. Repetition and consolidation of the material covered.

1. Answers to questions on homework (analysis of unsolved problems).

2. Monitoring the assimilation of the material (test).

III. Learning new material.

Solving complex inequalities with modules or parameters in them.

Let us solve the inequality |x – 1| < 3.

First, let's solve this inequality analytically by considering two cases:

a) If x – 1 > 0, i.e. x > 1, then |x – 1| = x – 1 and the inequality looks like x – 1< 3. Решение этого неравенства х < 4. Учитывая условие х >1, in this case we obtain solution 1< х < 4 или х [ 1; 4).

b) If x – 1< 0, т. е. х < 1, то |x – 1| = – (х – 1) = 1 – х и неравенство имеет вид 1 – х < 3. Решение этого неравенства -2 < х. Учитывая условие х < 1, получаем в этом случае решение -2 <х < 1 или х (-2; 1).

We find the union of the obtained solutions.

Since writing the answer in problems with parameters is very important (the answer is written in ascending order of the parameter), we give the full answer:

When a< 1 х [ а + 1; +); при а = 1 х (-; + ); при а >1 x (-; a + 1].

Now let's look at linear inequalities in two variables. As a rule, such problems are reduced to depicting a set of points whose coordinates satisfy the inequality on the coordinate plane.

On the coordinate plane we depict a set of points whose coordinates satisfy the inequality y-2 > x-3.

Let's write this inequality in the form y > x-1. First, let's plot the linear function y = x-1 (straight line). This line divides all points of the coordinate plane into points located on this line and points located under this line. Let's check which points satisfy this inequality.

From the first area we take, for example, control point A (0; 0) - the origin of coordinates. It is easy to check that then the inequality y > -1 holds. From the second area we select, for example, control point B (1; -1). For such a point the inequality y > x-1 does not hold. Consequently, this inequality is satisfied by points located above and on the line y = x-1 (i.e. points similar to point A). These points are shaded.

For what values ​​of the parameter a does the equation ax 2 + x – 1 = 0 have no solutions?

Since the leading coefficient of the equation depends on the parameter a, it is necessary to consider two cases.

a) If a 0, then the equation ax 2 + x – 1 = 0 is quadratic. Such an equation has no solutions if its discriminant D< 0. Решение этого неравенства а (-; -). Заметим, что в указанный промежуток значение а = 0 не входит.

b) If a = 0, then the equation ax 2 + x – 1 = 0 is linear and has the form x – 1 = 0. Obviously, the equation has a unique solution x = 1.

So, for a (-; -) this equation has no solutions.

Let us solve the inequality |x – 1| + x 2 + 2 x + 1< 0.

Let us write the inequality in the form |x – 1| + (x + 1) 2< 0 и введем новую переменную, а = х + 1. Тогда неравенство примет вид, |a| + а 2 < 0. Так как |a| >0 and a 2 > 0 for all values ​​of a, then the sum

|a| + a 2 > 0 for all a. Therefore the inequality, |a| + a 2< 0 имеет единственное решение а = 0. теперь вернемся к старой неизвестной х. Получаем линейное уравнение х + 1 = 0, решение которого х = – 1. Итак, решение данного неравенства х = – 1.

Similar types of inequalities exist with two variables.

On the coordinate plane we depict a set of points whose coordinates satisfy the inequality y-1< х 2 .

Let us write the inequality in the form y< х 2 + 1 и построим параболу y = х 2 + 1 (этот график получается смещением графика y = х 2 на одну единицу вверх). Парабола разбивает точки плоскости на точки, расположенные под параболой. Взяв в качестве контрольной точки начало координат, получаем верное неравенство 0 < 1. Поэтому данному неравенству удовлетворяют точки, расположенные ниже параболы и на параболе. Эти точки заштрихованы.

IV. Assignment in class and at home.

1. Solve the inequality analytically:

2. For all values ​​of a, solve the inequality:

3. At what values ​​of the parameter a does the equation

a) 3x 2 – 2x + a = 0 has no roots;
b) 2x 2 – 3x + 5a = 0 has two different roots;
c) 3akh 2 – 4х + 1 = 0 has two different roots;
d) ax 2 – 3x + 2 = 0 has at least one root.

4. Solve analytically (and if possible, graphically) the inequalities:

What you need to know about inequality icons? Inequalities with icon more (> ), or less (< ) are called strict. With icons more or equal (≥ ), less or equal (≤ ) are called not strict. Icon not equal (≠ ) stands apart, but you also have to solve examples with this icon all the time. And we will decide.)

The icon itself does not have much influence on the solution process. But at the end of the decision, when choosing the final answer, the meaning of the icon appears in full force! This is what we will see below in examples. There are some jokes there...

Inequalities, like equalities, exist faithful and unfaithful. Everything is simple here, no tricks. Let's say 5 > 2 is a true inequality. 5 < 2 - incorrect.

This preparation works for inequalities any kind and simple to the point of horror.) You just need to correctly perform two (only two!) elementary actions. These actions are familiar to everyone. But, characteristically, mistakes in these actions are the main mistake in solving inequalities, yes... Therefore, these actions must be repeated. These actions are called like this:

Identical transformations of inequalities.

Identical transformations of inequalities are very similar to identical transformations of equations. Actually, this is the main problem. The differences go over your head and... here you are.) Therefore, I will especially highlight these differences. So, the first identical transformation of inequalities:

1. The same number or expression can be added (subtracted) to both sides of the inequality. Any. This will not change the inequality sign.

In practice, this rule is used as a transfer of terms from the left side of the inequality to the right (and vice versa) with a change of sign. With a change in the sign of the term, not the inequality! The one-to-one rule is the same as the rule for equations. But the following identical transformations in inequalities differ significantly from those in equations. So I highlight them in red:

2. Both sides of the inequality can be multiplied (divided) by the same thingpositivenumber. For anypositive Will not change.

3. Both sides of the inequality can be multiplied (divided) by the same thingnegative number. For anynegativenumber. The inequality sign from thiswill change to the opposite.

You remember (I hope...) that the equation can be multiplied/divided by anything. And for any number, and for an expression with an X. If only it wasn't zero. This makes him, the equation, neither hot nor cold.) It does not change. But inequalities are more sensitive to multiplication/division.

A clear example for a long memory. Let us write an inequality that does not raise doubts:

5 > 2

Multiply both sides by +3, we get:

15 > 6

Any objections? There are no objections.) And if we multiply both sides of the original inequality by -3, we get:

15 > -6

And this is an outright lie.) A complete lie! Deception of the people! But as soon as you change the inequality sign to the opposite one, everything falls into place:

15 < -6

I’m not just swearing about lies and deception.) "Forgot to change the equal sign..."- This home error in solving inequalities. This trivial and simple rule has hurt so many people! Which they forgot...) So I’m swearing. Maybe I'll remember...)

Particularly attentive people will notice that inequality cannot be multiplied by an expression with an X. Respect to those who are attentive!) Why not? The answer is simple. We don’t know the sign of this expression with an X. It can be positive, negative... Therefore, we do not know which inequality sign to put after multiplication. Should I change it or not? Unknown. Of course, this restriction (the prohibition of multiplying/dividing an inequality by an expression with an x) can be circumvented. If you really need it. But this is a topic for other lessons.

That's all the identical transformations of inequalities. Let me remind you once again that they work for any inequalities Now you can move on to specific types.

Linear inequalities. Solution, examples.

Linear inequalities are inequalities in which x is in the first power and there is no division by x. Type:

x+3 > 5x-5

How are such inequalities resolved? They are very easy to solve! Namely: with the help of we reduce the most confusing linear inequality straight to the answer. That's the solution. I will highlight the main points of the decision. To avoid stupid mistakes.)

Let's solve this inequality:

x+3 > 5x-5

We solve it in exactly the same way as a linear equation. With the only difference:

We carefully monitor the inequality sign!

The first step is the most common one. With X's - to the left, without X's - to the right... This is the first identical transformation, simple and trouble-free.) Just don't forget to change the signs of the transferred terms.

The inequality sign remains:

x-5x > -5-3

Here are similar ones.

The inequality sign remains:

4x > -8

It remains to apply the last identical transformation: divide both sides by -4.

Divide by negative number.

The inequality sign will change to the opposite:

X < 2

This is the answer.

This is how all linear inequalities are solved.

Attention! Point 2 is drawn white, i.e. unpainted. Empty inside. This means that she is not included in the answer! I drew her so healthy on purpose. Such a point (empty, not healthy!)) in mathematics is called punctured point.

The remaining numbers on the axis can be marked, but not necessary. Extraneous numbers that are not related to our inequality can be confusing, yes... You just need to remember that the numbers increase in the direction of the arrow, i.e. numbers 3, 4, 5, etc. are to the right are twos, and numbers are 1, 0, -1, etc. - to the left.

Inequality x < 2 - strict. X is strictly less than two. If in doubt, checking is simple. We substitute the dubious number into the inequality and think: “Two is less than two? No, of course!” Exactly. Inequality 2 < 2 incorrect. A two in return is not appropriate.

Is one okay? Certainly. Less... And zero is good, and -17, and 0.34... Yes, all numbers that are less than two are good! And even 1.9999.... At least a little bit, but less!

So let's mark all these numbers on the number axis. How? There are options here. Option one is shading. We move the mouse over the picture (or touch the picture on the tablet) and see that the area of ​​​​all x's that meet the condition x is shaded < 2 . That's all.

Let's look at the second option using the second example:

X ≥ -0,5

Draw an axis and mark the number -0.5. Like this:

Notice the difference?) Well, yes, it’s hard not to notice... This dot is black! Painted over. This means -0.5 is included in the answer. Here, by the way, the verification may confuse someone. Let's substitute:

-0,5 ≥ -0,5

How so? -0.5 is no more than -0.5! And there is more icon...

It's OK. In a weak inequality, everything that fits the icon is suitable. AND equals good, and more good. Therefore, -0.5 is included in the response.

So, we marked -0.5 on the axis; it remains to mark all the numbers that are greater than -0.5. This time I mark the area of ​​suitable x values bow(from the word arc), rather than shading. We hover the cursor over the drawing and see this bow.

There is no particular difference between the shading and the arms. Do as the teacher says. If there is no teacher, draw arches. In more complex tasks, shading is less obvious. You can get confused.

This is how linear inequalities are drawn on an axis. Let us move on to the next feature of the inequalities.

Writing the answer for inequalities.

The equations were good.) We found x and wrote down the answer, for example: x=3. There are two forms of writing answers in inequalities. One is in the form of final inequality. Good for simple cases. For example:

X< 2.

This is a complete answer.

Sometimes you need to write down the same thing, but in a different form, at numerical intervals. Then the recording starts to look very scientific):

x ∈ (-∞; 2)

Under the icon ∈ the word is hidden "belongs".

The entry reads like this: x belongs to the interval from minus infinity to two not including. Quite logical. X can be any number from all possible numbers from minus infinity to two. There cannot be a double X, which is what the word tells us "not including".

And where in the answer is it clear that "not including"? This fact is noted in the answer round bracket immediately after the two. If the two were included, the bracket would be square. Like this one: ]. The following example uses such a parenthesis.

Let's write down the answer: x ≥ -0,5 at intervals:

x ∈ [-0.5; +∞)

Reads: x belongs to the interval from minus 0.5, including, to plus infinity.

Infinity can never be turned on. It's not a number, it's a symbol. Therefore, in such notations, infinity is always adjacent to a parenthesis.

This form of recording is convenient for complex answers consisting of several spaces. But - just for final answers. In intermediate results, where a further solution is expected, it is better to use the usual form, in the form of a simple inequality. We will deal with this in the relevant topics.

Popular tasks with inequalities.

The linear inequalities themselves are simple. Therefore, tasks often become more difficult. So it was necessary to think. This, if you’re not used to it, is not very pleasant.) But it’s useful. I will show examples of such tasks. Not for you to learn them, it's unnecessary. And in order not to be afraid when meeting such examples. Just think a little - and it’s simple!)

1. Find any two solutions to the inequality 3x - 3< 0

If it’s not very clear what to do, remember the main rule of mathematics:

If you don’t know what you need, do what you can!)

X < 1

And what? Nothing special. What are they asking us? We are asked to find two specific numbers that are the solution to an inequality. Those. fit the answer. Two any numbers. Actually, this is confusing.) A couple of 0 and 0.5 are suitable. A couple -3 and -8. There are an infinite number of these couples! Which answer is correct?!

I answer: everything! Any pair of numbers, each of which is less than one, will be the correct answer. Write which one you want. Let's move on.

2. Solve the inequality:

4x - 3 ≠ 0

Tasks in this form are rare. But, as auxiliary inequalities, when finding ODZ, for example, or when finding the domain of definition of a function, they occur quite often. Such a linear inequality can be solved as an ordinary linear equation. Only everywhere except the "=" sign ( equals) put a sign " ≠ " (not equal). This is how you approach the answer, with an inequality sign:

X ≠ 0,75

In more complex examples, it is better to do things differently. Make inequality out of equality. Like this:

4x - 3 = 0

Calmly solve it as taught and get the answer:

x = 0.75

The main thing is, at the very end, when writing down the final answer, do not forget that we found x, which gives equality. And we need - inequality. Therefore, we don’t really need this X.) And we need to write it down with the correct symbol:

X ≠ 0,75

This approach results in fewer errors. Those who solve equations automatically. And for those who don’t solve equations, inequalities are, in fact, of no use...) Another example of a popular task:

3. Find the smallest integer solution to the inequality:

3(x - 1) < 5x + 9

First we simply solve the inequality. We open the brackets, move them, bring similar ones... We get:

X > - 6

Didn't it work out that way!? Did you follow the signs!? And behind the signs of members, and behind the sign of inequality...

Let's think again. We need to find a specific number that matches both the answer and the condition "smallest integer". If it doesn’t dawn on you right away, you can just take any number and figure it out. Two over minus six? Certainly! Is there a suitable smaller number? Of course. For example, zero is greater than -6. And even less? We need the smallest thing possible! Minus three is more than minus six! You can already catch the pattern and stop stupidly going through the numbers, right?)

Let's take a number closer to -6. For example, -5. The answer is fulfilled, -5 > - 6. Is it possible to find another number less than -5 but greater than -6? You can, for example, -5.5... Stop! We are told whole solution! Doesn't roll -5.5! What about minus six? Uh-uh! The inequality is strict, minus 6 is in no way less than minus 6!

Therefore, the correct answer is -5.

I hope everything is clear with the choice of value from the general solution. Another example:

4. Solve inequality:

7 < 3x+1 < 13

Wow! This expression is called triple inequality. Strictly speaking, this is an abbreviated form of a system of inequalities. But such triple inequalities still have to be solved in some tasks... It can be solved without any systems. According to the same identical transformations.

We need to simplify, bring this inequality to pure X. But... What should be moved where?! This is where it’s time to remember that moving left and right is short form first identity transformation.

And the full form sounds like this: Any number or expression can be added/subtracted to both sides of the equation (inequality).

There are three parts here. So we will apply identical transformations to all three parts!

So, let's get rid of the one in the middle part of the inequality. Let's subtract one from the entire middle part. So that the inequality does not change, we subtract one from the remaining two parts. Like this:

7 -1< 3x+1-1 < 13-1

6 < 3x < 12

That’s better, right?) All that remains is to divide all three parts into three:

2 < X < 4

That's all. This is the answer. X can be any number from two (not including) to four (not including). This answer is also written down at intervals, such entries will be in quadratic inequalities. There they are the most common thing.

At the end of the lesson I will repeat the most important thing. Success in solving linear inequalities depends on the ability to transform and simplify linear equations. If at the same time watch for the inequality sign, there won't be any problems. That's what I wish for you. No problems.)

If you like this site...

By the way, I have a couple more interesting sites for you.)

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You can get acquainted with functions and derivatives.