The method of solving an indicative inequality with the smallest degree. Solution of indicative inequalities: basic ways

Theory:

When solving inequalities, the following rules use:

1. Any member of inequality can be transferred from one part.
inequalities to another with the opposite sign, while the sign of inequality does not change.

2. Both parts of inequality can be multiplied or divided into one
and the same positive number without changing the sign of inequality.

3. Both parts of inequality can be multiplied or divided into one
and also a negative numberby changing the sign of inequality on
opposite.

Solve inequality - 8 x + 11< − 3 x − 4
Decision.

1. We suffer a member - 3 X. into the left part of inequality and a member 11 - to the right side of inequality, while changing signs on opposite - 3 X. and u 11 .
Then we get

- 8 x + 3 x< − 4 − 11

- 5 X.< − 15

2. We split both parts of inequality - 5 X.< − 15 on a negative number − 5 , with the sign of inequality < , Changes on > . We will turn to the inequality of the opposite sense.
We get:

- 5 X.< − 15 | : (− 5 )

x\u003e - 15: (- 5)

x\u003e 3.

x\u003e 3. - Decision given inequality.

Pay attention!

To write a solution, you can use two options: x\u003e 3. or in the form of a numerical gap.

We note many solutions of inequality on a numerical direct and write the answer in the form of a numerical gap.

x ∈ (3 ; + ∞ )

Answer: x\u003e 3. or x ∈ (3 ; + ∞ )

Algebraic inequalities.

Square inequalities. Rational inequalities of the highest degrees.

Methods of solutions of inequalities depend mainly on how the functions include the functions of inequality.

I.. Square inequalities, that is, the inequalities of the form

aX 2 + BX + C\u003e 0 (< 0), a ≠ 0.

To solve inequality, you can:

Square threefold decompose on multipliers, that is, to record inequality in the form

a (x - x 1) (x - x 2)\u003e 0 (< 0).

The roots of the polynomial are applied to the numeric axis. The roots break the many valid numbers into the gaps, in each of which the corresponding quadratic function will be a sign resistant.
Determine the sign a (x - x 1) (x - x 2) in each interval and write down the answer.

If the square threestrate does not have roots, then when D<0 и a>0 Square threefold at any x is positive.

Solve inequality. x 2 + x - 6\u003e 0.

We decompose a square three-stake on multipliers (x + 3) (x - 2)\u003e 0

Answer: X (-∞; -3) (2; + ∞).

2) (x - 6) 2\u003e 0

This inequality is true for any x, except x \u003d 6.

Answer: (-∞; 6) (6; + ∞).

3) x² + 4x + 15< 0.

Here D.< 0, a = 1 > 0. Square threefold is positive for all x.

Answer: X î Ø.

Solve inequalities:

1 + x - 2xqm< 0. Ответ:
3xqm - 12x + 12 ≤ 0. Answer:
3xqm - 7x + 5 ≤ 0. Answer:
2xqm - 12x + 18\u003e 0. Answer:
Under what values \u200b\u200bare inequality

x² - AX\u003e performed for any x? Answer:

II.. Rational inequalities of the highest degrees, That is the inequalities of the form

a n x n + a n-1 x n-1 + ... + a 1 x + a 0\u003e 0 (<0), n>2.

The highest degree should be decisled on multipliers, that is, the inequality is recorded as

a n (x - x 1) (x - x 2) · ... · (x - x n)\u003e 0 (<0).

Mark on the numeric axis of the point in which the polynomial adds to zero.

Determine the signs of the polynomial at each interval.

1) Solve inequality x 4 - 6x 3 + 11x 2 - 6x< 0.

x 4 - 6x 3 + 11x 2 - 6x \u003d x (x 3 - 6x 2 + 11x -6) \u003d x (x 3 - x 2 - 5x 2 + 5x + 6x - 6) \u003d x (x - 1) (x 2 -5x + 6) \u003d

x (x - 1) (x - 2) (x - 3). So, x (x - 1) (x - 2) (x - 3)<0

Answer: (0; 1) (2; 3).

2) solve the inequality (x -1) 5 (x + 2) (x - ½) 7 (2x + 1) 4<0.

Note on the numeric axis of the point in which the polynomial adds to zero. It is x \u003d 1, x \u003d -2, x \u003d ½, x \u003d - ½.

At point x \u003d - ½, the sign of the sign does not occur, because bounce (2x + 1) is erected into an even degree, that is, the expression (2x + 1) 4 does not change the sign when switching through the point x \u003d - ½.

Answer: (-∞; -2) (½; 1).

3) Solve inequality: x 2 (x + 2) (x - 3) ≥ 0.

This inequality is equivalent to the following aggregate.

The solution (1) is x (-∞; -2) (3; + ∞). The solution (2) is x \u003d 0, x \u003d -2, x \u003d 3. Combining the solutions obtained, we obtain x î (-∞; -2] (0) (0)

Where the role of $ b $ may be a common number, and maybe something else. Examples? Yes please:

\\ [\\ begin (align) & ((2) ^ (x)) \\ gt 4; \\ quad ((2) ^ (x - 1)) \\ le \\ FRAC (1) (\\ SQRT (2)); \\ \\\\ & ((0.1) ^ (1-x)) \\ lt 0.01; \\ quad ((2) ^ (\\ FRAC (X) (2))) \\ lt ((4) ^ (\\ FRAC (4) (x))). \\\\\\ End (Align) \\]

I think the meaning is clear: there is an indicative function $ ((a) ^ (x)) $, compared with something, and then asked to find $ x $. In particular clinical cases, instead of a variable $ x $, some function $ F \\ left (X \\ Right) $ and thereby becoming a little bit inequality. :)

Of course, in some cases, inequality may look more severe. For example:

\\ [((9) ^ (x)) + 8 \\ gt ((3) ^ (x + 2)) \\]

Or even here:

In general, the complexity of such inequalities may be the most different, but in the end they are still reduced to a simple design of $ ((a) ^ (x)) \\ gt b $. And with such a design, we somehow understand (in particular clinical cases, when nothing comes to mind, logarithms will help us). Therefore, now we will mention such simple designs.

The solution of the simplest demonstration inequalities

Consider anything quite simple. For example, this is:

\\ [((2) ^ (x)) \\ gt 4 \\]

Obviously, the number of the right can be rewritten in the form of a degree of two: $ 4 \u003d ((2) ^ (2)) $. Thus, the initial inequality will rewrite in a very convenient form:

\\ [((2) ^ (x)) \\ gt ((2) ^ (2)) \\]

And now the hands will be scratching "cross" the twos standing in the grounds of degrees in order to get the answer $ X \\ GT $ 2. But before what to cheat there, let's remember the degrees:

\\ [((2) ^ (1)) \u003d 2; \\ quad (((2) ^ (2)) \u003d 4; \\ quad ((2) ^ (3)) \u003d 8; \\ quad ((2) ^ ( 4)) \u003d 16; ... \\]

As we see than more It is worth the extent, the more the number at the output is. "Thanks, Cap!" - exclaim someone from the disciples. Is it different? Unfortunately, it happens. For example:

\\ [((\\ left (\\ FRAC (1) (2) \\ RIGHT)) ^ (1)) \u003d \\ FRAC (1) (2); \\ quad ((\\ left (\\ FRAC (1) (2) \\ ); ... \\]

Here, too, everything is logical: the more degree, the more times the number 0.5 is multiplied by itself (that is, it is divided into half). Thus, the resulting sequence of numbers decreases, and the difference between the first and second sequence consists only at the base:

If the base of the degree $ A \\ gt $ 1, then as the indicator grows the $ n $ number $ ((a) ^ (n)) $ will also grow;
And on the contrary, if $ 0 \\ lt a \\ lt $ 1 $, as the indicator grows the $ n $ number $ ((a) ^ (n)) $ will decrease.

Summing these facts, we get the most important statement on which all the solution of indicative inequalities is founded:

If $ A \\ GT $ 1, then the inequality $ ((a) ^ (x)) \\ gt ((a) ^ (n)) $ is equivalent to the inequality of $ x \\ gt n $. If $ 0 \\ lt a \\ lt 1 $, then the inequality is $ ((a) ^ (x)) \\ gt ((a) ^ (n)) $ is equivalent to the inequality of $ x \\ lt n $.

In other words, if the basis is greater than the unit, it can be simply removed - the sign of inequality will not change. And if the base is less than one, it can also be removed, but at the same time you have to change the sign of inequality.

Please note: we did not consider the options $ a \u003d 1 $ and $ A \\ Le 0 $. Because in these cases uncertainty arises. Suppose how to solve the inequality of the type $ ((1) ^ (x)) \\ gt $ 3? The unit to any extent will give a unit again - we will never get a triple or more. Those. There are no solutions.

With negative grounds, it is still more interesting. Consider for the example here such inequality:

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

At first glance, everything is simple:

Right? And here is not! It is enough to substitute instead of $ x $ a couple of ready and a couple of odd numbers to make sure that the decision is incorrect. Take a look:

\\ [\\ begin (align) & x \u003d 4 \\ rightarrow ((\\ left (-2 \\ right)) ^ (4)) \u003d 16 \\ gt 4; \\\\ & x \u003d 5 \\ rightarrow ((\\ left (-2 \\ right)) ^ (5)) \u003d - 32 \\ lt 4; \\\\ & x \u003d 6 \\ rightarrow ((\\ left (-2 \\ right)) ^ (6)) \u003d 64 \\ gt 4; \\\\ & x \u003d 7 \\ rightarrow ((\\ left (-2 \\ right)) ^ (7)) \u003d - 128 \\ lt 4. \\\\\\ End (Align) \\]

As you can see, signs alternate. But there is still fractional degrees And so on tin. How, for example, order to count $ ((\\ left (-2 \\ right)) ^ (\\ sqrt (7))) $ (minus twice to the degree of the root of seven)? Yes, nothing!

Therefore, for certainty, it is believed that in all indicative inequalities (and equations, by the way, too) $ 1 \\ ne a \\ gt 0 $. And then everything is solved very simply:

\\ [((a) ^ (x)) \\ gt ((a) ^ (n)) \\ rightarrow \\ left [\\ begin (align) & x \\ gt n \\ quad \\ left (a \\ gt 1 \\ right), \\\\ & x \\ lt n \\ quad \\ left (0 \\ lt a \\ lt 1 \\ right). \\\\\\ End (Align) \\ Right. \\]

In general, once again remember the main rule: if the basis in the indicative equation is greater than one, it can be simply removed; And if the base is less than one, it can also be removed, but the sign of inequality will change.

Examples of solutions

So, consider some simple demonstration inequalities:

\\ [\\ begin (Align) & ((2) ^ (x - 1)) \\ le \\ FRAC (1) (\\ SQRT (2)); \\\\ & ((0.1) ^ (1-x)) \\ LT 0.01; \\\\ & ((2) ^ (((x) ^ (2)) - 7x + 14)) \\ LT 16; \\\\ & ((0.2) ^ (1 + ((x) ^ (2)))) \\ GE \\ FRAC (1) (25). \\\\\\ End (Align) \\]

The primary task in all cases is the same: to reduce inequalities to the simplest type of $ ((a) ^ (x)) \\ gt ((a) ^ (n)) $. That is what we now make with each inequality, and at the same time we repeat the properties of the degrees and the indicative function. So let's go!

\\ [((2) ^ (x - 1)) \\ le \\ FRAC (1) (\\ SQRT (2)) \\]

What can I do here? Well, on the left, we have a demonstration expression, it is not necessary to change anything. But on the right there is some kind of crap: the fraction, and even in the denominator root!

However, let us remember the rules for working with fractions and degrees:

\\ [\\ begin (Align) & \\ FRAC (1) (((a) ^ (n))) \u003d ((a) ^ (- n)); \\\\ \\ sqrt [k] (a) \u003d ((a) ^ (\\ FRAC (1) (K))). \\\\\\ End (Align) \\]

What does it mean? First, we can easily get rid of the fraction, turning it into a degree with a negative indicator. And secondly, since the root stands in the denominator, it would be nice to turn it into a degree - this time with a fractional indicator.

Apply these actions consistently to the right side of inequality and see what happens:

\\ [\\ FRAC (1) (\\ SQRT (2)) \u003d ((\\ left (\\ sqrt (2) \\ right)) ^ (- 1)) \u003d ((\\ left (((2) ^ (\\ FRAC ( 1) (3))) \\ Right)) ^ (- 1)) \u003d ((2) ^ (\\ FRAC (1) (3) \\ CDOT \\ LEFT (-1 \\ RIGHT))) \u003d ((2) ^ (- \\ FRAC (1) (3))) \\]

Do not forget that when erecting the degree in the degree of indicators of these degrees add up. In general, when working with indicative equations and inequalities, it is absolutely necessary to know at least the simplest rules for working with degrees:

\\ [\\ begin (align) & ((a) ^ (x)) \\ Cdot ((a) ^ (y)) \u003d ((a) ^ (x + y)); \\\\ \\ FRAC (((a) ^ (x))) (((a) ^ (y))) \u003d ((a) ^ (x-y)); \\\\ & ((\\ left ((((a) ^ (x)) \\ Right)) ^ (y)) \u003d ((a) ^ (x \\ cdot y)). \\\\\\ End (Align) \\]

Actually, the last rule we just applied. Therefore, our initial inequality will be rewriting as follows:

\\ [((2) ^ (x - 1)) \\ le \\ FRAC (1) (\\ sqrt (2)) \\ rightarrow ((2) ^ (x - 1)) \\ le ((2) ^ (- \\ Now get rid of two at the base. Since 2\u003e 1, the inequality sign will remain the same:

\\ [\\ begin (align) & x-1 \\ le - \\ FRAC (1) (3) \\ rightarrow x \\ le 1- \\ FRAC (1) (3) \u003d \\ FRAC (2) (3); \\\\ & X \\ IN \\ LEFT (- \\ INFTY; \\ FRAC (2) (3) \\ RIGHT]. \\\\\\ End (Align) \\]

That's all the decision! The main difficulty is not at all in the impact function, but in the competent transformation of the original expression: you need to carefully and maximize it to bring it to the simplest mind.

Consider the second inequality:

\\ [((0.1) ^ (1-x)) \\ lt 0.01 \\]

So-so. Here we will wait for decimal fractions. As I have already spoken many times, in any expressions with degrees, you should get rid of decimal fractions - it is often possible to see a quick and simple solution. So we will get rid of:

\\ [\\ begin (align) & 0.1 \u003d \\ FRAC (1) (10); \\ quad 0,01 \u003d \\ FRAC (1) (100) \u003d ((\\ left (\\ FRAC (1) (10) \\ \\\\ & ((0.1) ^ (1-x)) \\ LT 0.01 \\ RIGHTARROW ((\\ left (\\ FRAC (1) (10) \\ RIGHT)) ^ (1-x)) \\ lt ( (\\ left (\\ FRAC (1) (10) \\ RIGHT)) ^ (2)). \\\\\\ End (Align) \\]

We have recently the simplest inequality, and even with the basis of 1/10, i.e. Little units. Well, we remove the foundations, passing the sign with the "less" to "more", and we get:

\\ [\\ begin (align) & 1-x \\ gt 2; \\\\ & -x \\ gt 2-1; \\\\ & -x \\ gt 1; \\\\ & x \\ lt -1. \\\\\\ End (Align) \\]

{!LANG-2921f20f4c2c371099327f5b5124316c!}

Received the final answer: $ x \\ in \\ left (- \\ infty; -1 \\ right) $. Please note: the answer is precisely a set, and in no case the design of the $ X \\ lt -1 $. Because formally such a design is not a lot, and inequality relative to the $ x $ variable. Yes, it is very simple, but this is not the answer!

Important note. This inequality could be solved in a different way - by bringing both parts to the degree with the basis, the large unit. Take a look:

\\ [\\ FRAC (1) (10) \u003d (((10) ^ (- 1)) \\ rightarrow ((\\ left (((10) ^ (- 1)) \\ Right)) ^ (1-x)) \\ ((10) ^ (- 1 \\ Cdot 2)) \\]

After such a transformation, we again obtain a demonstrative inequality, but with the base 10\u003e 1. And this means that you can simply cross the top ten - the sign of inequality will not change. We get:

\\ [\\ begin (align) & -1 \\ cdot \\ left (1-x \\ right) \\ lt -1 \\ cdot 2; \\\\ & x-1 \\ lt -2; \\\\ & x \\ lt -2 + 1 \u003d -1; \\\\ & x \\ lt -1. \\\\\\ End (Align) \\]

As you can see, the answer turned out to be the same. At the same time, we saved yourself from the need to change the sign and generally remember some rules there. :)

\\ [((2) ^ (((x) ^ (2)) - 7x + 14)) \\ lt 16 \\]

However, let it be frightened. So that neither is in the indicators, the technology of solving the inequality itself remains the same. Therefore, we note to begin with that 16 \u003d 2 4. I rewrite the original inequality, taking into account this fact:

\\ [\\ begin (align) & ((2) ^ (((x) ^ (2)) - 7x + 14)) \\ lt ((2) ^ (4)); \\\\ & ((x) ^ (2)) - 7x + 14 \\ LT 4; \\\\ & ((x) ^ (2)) - 7x + 10 \\ LT 0. \\\\\\ End (Align) \\]

Hooray! We got the usual square inequality! The sign has not changed anywhere, because at the base there is a twice - a number, more units.

Zero functions on a numeric direct

We set the signs of the function $ f \\ left (x \\ right) \u003d ((x) ^ (2)) - 7x + $ 10 - obviously, it will be a parabol with branches up, so there will be "pluses". We are interested in the area where the function is less than zero, i.e. $ X \\ in \\ Left (2; 5 \\ Right) $ is the answer to the initial task.

Finally, consider another inequality:

\\ [((0.2) ^ (1 + ((x) ^ (2)))) \\ GE \\ FRAC (1) (25) \\]

Again you see indicative function With a decimal fraction at the base. Transfer this fraction to ordinary:

\\ [\\ Begin (Align) & 0.2 \u003d \\ FRAC (2) (10) \u003d \\ FRAC (1) (5) \u003d ((5) ^ (- 1)) \\ Rightarrow \\\\ & \\ Rightarrow ((0 , 2) ^ (1 + ((x) ^ (2)))) \u003d ((\\ left ((((5) ^ (- 1)) \\ right)) ^ (1 + (((x) ^ (2) ))) \u003d ((5) ^ (- 1 \\ Cdot \\ left (1 + ((x) ^ (2)) \\ Right))) \\ End (Align) \\]

In this case, we took advantage of the previously given remarks - the base was reduced to the number 5\u003e 1 to simplify the further decision. In the same way and with the right part:

\\ [\\ FRAC (1) (25) \u003d ((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (2)) \u003d ((\\ left (((5) ^ (- 1)) \\ I rewrite the original inequality, taking into account both transformations:

\\ [(((0.2) ^ (1 + ((x) ^ (2)))) \\ GE \\ FRAC (1) (25) \\ rightarrow ((5) ^ (- 1 \\ Cdot \\ left (1+ ((x) ^ (2)) \\ RIGHT))) \\ GE ((5) ^ (- 2)) \\]

The bases on both sides are the same and superior to one. There are no other terms on the right and on the left, so they simply "tighten" five and get a simple expression:

\\ [\\ begin (align) & -1 \\ cdot \\ left (1 + ((x) ^ (2)) \\ Right) \\ GE -2; \\\\ & -1 - ((x) ^ (2)) \\ GE -2; \\\\ & - ((x) ^ (2)) \\ GE -2 + 1; \\\\ & - ((x) ^ (2)) \\ GE -1; \\ quad \\ left | \\ Cdot \\ Left (-1 \\ Right) \\ Right. \\\\ & ((x) ^ (2)) \\ LE 1. \\\\\\ End (Align) \\]

Here it is necessary to be careful. Many students love just extract

Square root their both parts of the inequality and write something in the spirit of $ x \\ le 1 \\ rightarrow x \\ in \\ left (- \\ infty; -1 \\ right] $. To do this in no case, because the root of the exact square is Module, and in no case is not the original variable: \\ [\\ SQRT (((x) ^ (2))) \u003d \\ left | X \\ Right | \\]

However, working with modules is not the most pleasant occupation, right? So we will not work. And instead, we simply move all the terms to the left and solve the usual inequality of intervals:

$ \\ begin (align) & ((x) ^ (2)) - 1 \\ le 0; \\\\ \\ left (x-1 \\ right) \\ left (x + 1 \\ right) \\ le 0 \\\\ \\ ((x) _ (1)) \u003d 1; \\ quad ((x) _ (2)) \u003d -1; \\\\\\ End (Align) $

We again note the points obtained on the numeric straight and watch the signs:

Note: Points are painted

Since we solved the incredible inequality, all points on the graph are painted. Therefore, the answer will be like this: $ x \\ in \\ left [-1; 1 \\ right] $ - not the interval, namely the segment.

In general, I would like to notice that there is nothing complicated in the indicative inequalities. The meaning of all the transformations that we were performed today are reduced to a simple algorithm:

Find the base to which we will solve all degrees;

Gently perform the conversion so that the inequality of the type $ ((a) ^ (x)) \\ gt ((a) ^ (n)) $ is obtained. Of course, instead of variables $ x $ and $ n $ can stand much more
Complex functions but the meaning of this will not change;Strike the foundation of degrees. At the same time, the sign of inequality may change if the base is $ a \\ lt $ 1.
In essence, this

Universal algorithm {!LANG-1e90d168ff8aef74d56da63cba48fbd8!} Solutions of all such inequalities. And all you will tell you on this topic - only specific techniques and tricks, allowing to simplify and speed up the transformation. Here we will talk about one of these techniques. :)

Rationalization method

Consider another batch of inequalities:

\\ [\\ begin (Align) & ((\\ Text () \\! \\! \\ pi \\! \\! \\ text ()) ^ (x + 7)) \\ gt ((\\ Text () \\! \\! \\ PI \\! \\! \\ text ()) ^ (((x) ^ (2)) - 3x + 2)); \\\\ & ((\\ left (2 \\ sqrt (3) -3 \\ right)) ^ (((x) ^ (2)) - 2x)) \\ lt 1; \\\\ & ((\\ left (\\ FRAC (1) (3) \\ RIGHT)) ^ (((x) ^ (2)) + 2x)) \\ gt ((\\ left (\\ FRAC (1) (9) \\ Right)) ^ (16-x)); \\\\ & ((\\ left (3-2 \\ SQRT (2) \\ RIGHT)) ^ (3x - ((x) ^ (2)))) \\ lt 1. \\\\\\ End (Align) \\]

So what's so special about them? They are lungs. Although, stop! The number π is erected into any degree? What nonsense?

And how to build a number of $ 2 \\ SQRT (3) -3 $? Or $ 3-2 \\ SQRT (2) $? The challenges obviously fought "hawthorn" before sitting for work. :)

In fact, nothing terrible in these tasks. Let me remind you: an indicative function is called the expression of the type $ ((a) ^ (x)) $, where the base $ a $ is any positive number, except for the unit. The number π positively is we also know. Numbers $ 2 \\ SQRT (3) -3 $ and $ 3-2 \\ SQRT (2) $ are also positive - it is easy to make sure if you compare them with zero.

It turns out that all these "frightening" inequalities do not differ from the simple, discussed above? And are solved in the same way? Yes, quite right. However, on their example, I would like to consider one reception that saves time to independent work and exams. It will be about the rationalization method. So, attention:

Any indicative inequality of the type $ ((a) ^ (x)) \\ gt ((a) ^ (n)) $ is equivalent to the inequality of $ \\ left (xn \\ right) \\ Cdot \\ left (A-1 \\ Right) \\ gt 0 $.

That's the whole method. :) And you thought that there would be some other game? Nothing like this! But this simple fact recorded literally in one line will greatly simplify us work. Take a look:

\\ [\\ begin (Matrix) ((\\ Text () \\! \\! \\ pi \\! \\! \\ text ()) ^ (x + 7)) \\ gt ((\\ text () \\! \\! \\ pi \\ -3x + 2 \\ Right) \\ Right) \\ Cdot \\ Left (\\ Text () \\! \\! \\ PI \\! \\! \\ Text () -1 \\ Right) \\ GT 0 \\\\\\ End (Matrix) \\]

So no more indicative functions! And do not remember: the sign changes or not. But arises new problem: What to do with the fell factor \\ [\\ left (\\ text () \\! \\! \\ PI \\! \\! \\ Text () -1 \\ Right) \\]? We do not know what is equal to the exact value of the number π. However, the captain is obvious as it hints:

\\ [\\ text () \\! \\! \\ pi \\! \\! \\ text () \\ approx 3,14 ... \\ gt 3 \\ rightarrow \\ text () \\! \\! \\ PI \\! \\! \\ TEXT ( ) -1 \\ gt 3-1 \u003d 2 \\]

In general, the exact value of π is particularly different, and not whether it is not whether it is only important for us to understand that in any case $ \\ text () \\! \\! \\ Pi \\! \\! \\ Text () -1 \\ gt $ 2, t .. This is a positive constant, and we can divide both part of inequality on it:

\\ [\\ begin (align) \\ left (x + 7- \\ left (((x) ^ (2)) - 3x + 2 \\ Right) \\ Right) \\ Cdot \\ Left (\\ Text () \\! \\! \\ PI \\! \\! \\ Text () -1 \\ Right) \\ gt 0 \\\\ & x + 7- \\ left (((x) ^ (2)) - 3x + 2 \\ Right) \\ GT 0; \\\\ & x + 7 - ((x) ^ (2)) + 3x-2 \\ gt 0; \\\\ & - ((x) ^ (2)) + 4x + 5 \\ gt 0; \\ quad \\ left | \\ Cdot \\ Left (-1 \\ Right) \\ Right. \\\\ & ((x) ^ (2)) - 4x-5 \\ lt 0; \\\\ & \\ Left (X-5 \\ Right) \\ Left (X + 1 \\ RIGHT) \\ LT 0. \\\\\\ End (Align) \\]

As you can see, at a certain point I had to divide the unit at the minus - at the same time the sign of inequality was changed. In the end, I decomposed the square triple on the Vieta theorem - it is obvious that the roots are equal to $ ((x) _ (1)) \u003d 5 $ and $ ((x) _ (2)) \u003d - $ 1. Further everything is solved by a classic interval method:

Solve the inequality by intervals

All points are inqualing, since the initial inequality is strict. We are interested in an area with negative values, so the answer is: $ x \\ in \\ left (-1; 5 \\ right) $. That's the whole solution. :)

Let us turn to the next task:

\\ [((\\ left (2 \\ sqrt (3) -3 \\ right)) ^ (((x) ^ (2)) - 2x)) \\ lt 1 \\]

Here, everything is simple, because the right is worth the unit. And we remember that the unit is any number to zero. Even if this number is an irrational expression, standing at the bottom of the left:

\\ [\\ begin (align) & ((\\ left (2 \\ sqrt (3) -3 \\ right)) ^ ((((x) ^ (2)) - 2x)) \\ lt 1 \u003d ((\\ left (2 \\ SQRT (3) -3 \\ RIGHT)) ^ (0)); \\\\ \\ (\\ left (2 \\ sqrt (3) -3 \\ right)) ^ (((x) ^ (2)) - 2x)) \\ lt ((\\ left (2 \\ sqrt (3) -3 \\ RIGHT)) ^ (0)); \\\\\\ End (Align) \\]

Well, we carry out rationalization:

\\ [\\ begin (align) \\ left (((x) ^ (2)) - 2x-0 \\ right) \\ Cdot \\ left (2 \\ SQRT (3) -3-1 \\ RIGHT) \\ LT 0; \\\\ \\ \\ left (((x) ^ (2)) - 2x-0 \\ Right) \\ CDOT \\ LEFT (2 \\ SQRT (3) -4 \\ RIGHT) \\ LT 0; \\\\ \\ left (((x) ^ (2)) - 2x-0 \\ Right) \\ CDot 2 \\ Left (\\ SQRT (3) -2 \\ RIGHT) \\ LT 0. \\\\\\ End (Align) \\ It remains only to deal with signs. The $ 2 \\ Left multiplier (\\ SQRT (3) -2 \\ RIGHT) $ does not contain a variable $ x $ is just a constant, and we need to figure out its sign. To do this, we note the following:

\\ [\\ begin (Matrix) \\ SQRT (3) \\ LT \\ SQRT (4) \u003d 2 \\\\ \\ DOWNARROW \\\\ 2 \\ Left (\\ SQRT (3) -2 \\ RIGHT) \\ LT 2 \\ CDOT \\ LEFT (2 -2 \\ RIGHT) \u003d 0 \\\\\\ End (Matrix) \\]

It turns out that the second factor is not just a constant, but a negative constant! And when dividing on it, the sign of the initial inequality will change to the opposite:

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\\ [\\ begin (align) \\ left (((x) ^ (2)) - 2x-0 \\ right) \\ CDOT 2 \\ left (\\ SQRT (3) -2 \\ RIGHT) \\ LT 0; \\\\ & ((x) ^ (2)) - 2x-0 \\ gt 0; \\\\ & X \\ Left (X-2 \\ Right) \\ GT 0. \\\\\\ End (Align) \\]

Now everything becomes completely obvious. The roots of the square triple, standing on the right: $ ((x) _ (1)) \u003d 0 $ and $ ((x) _ (2)) \u003d $ 2. We note them on a numeric straight and see the signs of the function $ f \\ left (x \\ right) \u003d x \\ left (x-2 \\ right) $:

The case is when we are interested in side intervals

We are interested in the intervals marked by the "Plus" sign. It only remains to write out the answer:

Go to the following example:

\\ [((\\ left (\\ FRAC (1) (3) \\ RIGHT)) ^ (((x) ^ (2)) + 2x)) \\ gt ((\\ left (\\ FRAC (1) (9) \\ Well, everything is completely obvious here: in the grounds there are the degrees of the same number. Therefore, I will write everything briefly:

\\ [\\ begin (Matrix) \\ FRAC (1) (3) \u003d ((3) ^ (- 1)); \\ quad \\ FRAC (1) (9) \u003d \\ FRAC (1) (((3) ^ ( 2))) \u003d ((3) ^ (- 2)) \\\\ \\ downarrow \\\\ ((\\ left (((3) ^ (- 1)) \\ right)) ^ (((x) ^ (2) ) + 2x)) \\ gt ((\\ left (((3) ^ (- 2)) \\ right)) ^ (16-x)) \\\\\\ End (Matrix) \\]

\\ [\\ begin (align) & ((3) ^ (- 1 \\ cdot \\ left (((x) ^ (2)) + 2x \\ right))) \\ gt ((3) ^ (- 2 \\ Cdot \\ \\\\ & ((3) ^ (- ((x) ^ (2)) - 2x)) \\ gt ((3) ^ (- 32 + 2x)); \\\\ \\ left (- ((x) ^ (2)) - 2x- \\ left (-32 + 2x \\ Right) \\ Right) \\ CDOT \\ LEFT (3-1 \\ RIGHT) \\ GT 0; \\\\ & - ((x) ^ (2)) - 2x + 32-2x \\ gt 0; \\\\ & - ((x) ^ (2)) - 4x + 32 \\ gt 0; \\ quad \\ left | \\ Cdot \\ Left (-1 \\ Right) \\ Right. \\\\ & ((x) ^ (2)) + 4x-32 \\ lt 0; \\\\ & \\ Left (X + 8 \\ RIGHT) \\ LEFT (X-4 \\ RIGHT) \\ LT 0. \\\\\\ End (Align) \\]

As you can see, in the process of transformations I had to multiply the negative number, so the sign of inequality was changed. At the very end, I re-applied the theorem of the Vieta to decompose on the multipliers of the square triple. As a result, the answer will be the following: $ X \\ in \\ left (-8; 4 \\ Right) $ - those who want to make sure that by drawing a numerical direct, noting the point and counting signs. And in the meantime, we will turn to the last inequality from our "set":

\\ [((\\ left (3-2 \\ sqrt (2) \\ right)) ^ (3x - ((x) ^ (2)))) \\ lt 1 \\]

As you can see, at the bottom there is again

irrational number and on the right again there is a unit. Therefore, we rewrite our indicative inequality as follows:\\ [((\\ left (3-2 \\ sqrt (2) \\ Right)) ^ (3x - ((x) ^ (2)))) \\ lt ((\\ left (3-2 \\ SQRT (2) \\ We apply rationalization:

\\ [\\ begin (align) \\ left (3x - ((x) ^ (2)) - 0 \\ right) \\ CDOT \\ left (3-2 \\ SQRT (2) -1 \\ RIGHT) \\ LT 0; \\\\ \\ left (3x - ((x) ^ (2)) - 0 \\ Right) \\ CDOT \\ LEFT (2-2 \\ SQRT (2) \\ RIGHT) \\ LT 0; \\\\ \\ left (3x - ((x) ^ (2)) - 0 \\ Right) \\ CDOT 2 \\ Left (1- \\ sqrt (2) \\ Right) \\ LT 0. \\\\\\ End (Align) \\

{!LANG-c58709e59490a17e4bd80811e928f0fd!}

{!LANG-a2d39fc15ed7cd452464f1d697e6f5af!}

However, it is quite obvious that $ 1- \\ sqrt (2) \\ lt 0 $, since $ \\ sqrt (2) \\ approx 1.4 ... \\ gt $ 1. Therefore, the second factor is a newly negative constant for which both part of inequality can be divided:

\\ [\\ begin (Matrix) \\ Left (3x - ((x) ^ (2)) - 0 \\ Right) \\ CDOT 2 \\ Left (1- \\ SQRT (2) \\ RIGHT) \\ LT 0 \\\\ \\ DOWNARROW \\ \\ [\\ begin (align) & 3x - ((x) ^ (2)) - 0 \\ gt 0; \\\\ & 3X - ((x) ^ (2)) \\ gt 0; \\ quad \\ left | \\ Cdot \\ Left (-1 \\ Right) \\ Right. \\\\ & ((x) ^ (2)) - 3x \\ lt 0; \\\\ & X \\ Left (X-3 \\ Right) \\ LT 0. \\\\\\ End (Align) \\]

Transition to another base

A separate problem in solving indicative inequalities is the search for "correct" base. Unfortunately, not always when I first look at the task, it is obvious that to take for the foundation, and what to do the degree of this foundation.

But do not worry: there is no magic and "secret" technologies. In mathematics, any skill, which cannot be algorithmized, can be easily developed by practice. But for this will have to solve problems

of different levels difficulties. For example, here are: \\ [\\ begin (Align) & ((2) ^ (\\ FRAC (X) (2))) \\ lt ((4) ^ (\\ FRAC (4) (X))); \\\\ & ((\\ left (\\ FRAC (1) (3) \\ RIGHT)) ^ (\\ FRAC (3) (x))) \\ GE ((3) ^ (2 + x)); \\\\ & ((\\ left (0.16 \\ RIGHT)) ^ (1 + 2x)) \\ CDOT ((\\ left (6.25 \\ RIGHT)) ^ (x)) \\ GE 1; \\\\ & ((\\ left (\\ FRAC (27) (\\ SQRT (3)) \\ Right)) ^ (- x)) \\ lt ((9) ^ (4-2x)) \\ CDOT 81. \\\\\\ Complicated? Scary? Yes, it is easier than the chicken on the asphalt! Let's try. First inequality:

\\ [((2) ^ (\\ FRAC (X) (2))) \\ lt ((4) ^ (\\ FRAC (4) (X))) \\]

Well, I think here everything is clear here:

We rewrite the original inequality, reducing everything to the "two" base:

\\ [((2) ^ (\\ FRAC (X) (2))) \\ lt ((2) ^ (\\ FRAC (8) (X))) \\ Rightarrow \\ left (\\ FRAC (X) (2) - \\ FRAC (8) (X) \\ RIGHT) \\ CDOT \\ LEFT (2-1 \\ RIGHT) \\ LT 0 \\]

Yes, yes, you all understood everything: I just applied the rationalization method described above. Now you need to work carefully: we had a fractional rational inequality (this is such a variable in the denominator), therefore, before equating something to zero, it is necessary to bring everything to the general denominator and get rid of the constant multiplier.

\\ [\\ begin (Align) & \\ Left (\\ FRAC (X) (2) - \\ FRAC (8) (X) \\ RIGHT) \\ Cdot \\ left (2-1 \\ Right) \\ LT 0; \\\\ \\ \\ left (\\ frac (((x) ^ (2)) - 16) (2x) \\ Right) \\ CDOT 1 \\ LT 0; \\\\ \\ FRAC (((x) ^ (2)) - 16) (2x) \\ lt 0. \\\\\\ End (Align) \\]

Now use

Standard method

Intervals. Nutrifier zeros: $ x \u003d \\ pm $ 4. The denominator refers to zero only at $ x \u003d 0 $. Total three points that should be noted on the numerical straight line (all points of otkoloty, because the sign of inequality is strict). We get: more Difficult case

{!LANG-d2d40c6211a5e644e371b36632756b9c!} {!LANG-80c638b121d35b095bdbc474902f5c78!}: Three roots

As it is not difficult to guess, the hatching is noted those intervals on which the expression on the left takes negative values. Therefore, the final response will go at once two intervals:

End of intervals are not in response, since the initial inequality was strict. No additional checks are required of this response. In this regard, the indicative inequalities are much easier than logarithmic: there are no restrictions, etc.

Go to the next task:

\\ [((\\ left (\\ FRAC (1) (3) \\ RIGHT)) ^ (\\ FRAC (3) (x))) \\ GE ((3) ^ (2 + x)) \\]

Here, too, no problems, since we already know that $ \\ FRAC (1) (3) \u003d (((3) ^ (- 1)) $, so all inequality can be rewritten so:

\\ [\\ begin (align) & ((\\ left (((3) ^ (- 1)) \\ Right)) ^ (\\ FRAC (3) (x))) \\ GE ((3) ^ (2 + x )) \\ Rightarrow ((3) ^ (- \\ FRAC (3) (x))) \\ ge ((3) ^ (2 + x)); \\\\ \\ \\ left (- \\ FRAC (3) (X) - \\ Left (2 + x \\ Right) \\ Right) \\ Cdot \\ Left (3-1 \\ Right) \\ GE 0; \\\\ & \\ LEFT (- \\ FRAC (3) (X) -2-X \\ RIGHT) \\ CDOT 2 \\ GE 0; \\ Quad \\ Left | : \\ Left (-2 \\ Right) \\ Right. \\\\ \\ FRAC (3) (X) + 2 + X \\ LE 0; \\\\ \\ FRAC (((x) ^ (2)) + 2x + 3) (x) \\ le 0. \\\\\\ End (Align) \\]

Please note: in the third line, I decided not to fine and immediately divide everything on (-2). The first bracket passed (now there are pluses everywhere), and the two decreased with the constant factor. This is how it is necessary to do when making real calculations for independent and test work - It is not necessary to paint every action and transformation.

Further, the interval method familiar to us is entering. Zeros of the numeral: and they are not. Because the discriminant will be negative. In turn, the denominator is reset only at $ x \u003d 0 $ - as the last time. Well, it is clear that on the right of $ x \u003d 0 $ the fraction will take positive meanings, and on the left are negative. Since we are interested precisely negative values, then the final answer: $ x \\ in \\ left (- \\ infty; 0 \\ right) $.

\\ [((\\ left (0.16 \\ RIGHT)) ^ (1 + 2x)) \\ CDOT ((\\ left (6.25 \\ RIGHT)) ^ (x)) \\ GE 1 \\]

And what should be done with decimal fractions in indicative inequalities? Right: get rid of them, translating into ordinary. So we will transfer:

\\ [\\ begin (align) & 0.16 \u003d \\ FRAC (16) (100) \u003d \\ FRAC (4) (25) \\ RIGHTARROW ((\\ left (0.16 \\ RIGHT)) ^ (1 + 2x)) \u003d ((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (1 + 2x)); \\\\ & 6,25 \u003d \\ FRAC (625) (100) \u003d \\ FRAC (25) (4) \\ RIGHTARROW ((\\ LEFT (6.25 \\ RIGHT)) ^ (x)) \u003d ((\\ left (\\ \\\\\\ End (Align) \\]

So what did we get in the grounds of the indicative functions? And we received two mutually reverse numbers:

\\ [\\ FRAC (25) (4) \u003d ((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (- 1)) \\ Rightarrow ((\\ Left (\\ FRAC (25) (4) \\ Left (\\ FRAC (4) (25) \\ RIGHT)) ^ (- x)) \\]

Thus, the initial inequality can be rewritten so:

\\ [\\ Begin (Align) & ((\\ Left (\\ FRAC (4) (25) \\ RIGHT)) ^ (1 + 2x)) \\ CDOT ((\\ Left (\\ FRAC (4) (25) \\ RIGHT) ) ^ (- x)) \\ GE 1; \\\\ & ((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (1 + 2x + \\ left (-x \\ right))) \\ GE ((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (0)); \\\\ \\ (\\ left (\\ FRAC (4) (25) \\ Right)) ^ (x + 1)) \\ ge ((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (0) ). \\\\\\ End (Align) \\]

Of course, when multiplying degrees with the same base, their indicators are folded, which happened in the second line. In addition, we presented a unit standing on the right, also in the form of a degree based on 4/25. It remains only to fulfill rationalization:

\\ [((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (x + 1)) \\ ge ((\\ left (\\ FRAC (4) (25) \\ RIGHT)) ^ (0)) \\ Rightarrow \\ left (X + 1-0 \\ RIGHT) \\ CDOT \\ LEFT (\\ FRAC (4) (25) -1 \\ Right) \\ GE 0 \\]

Note that $ \\ FRAC (4) (25) -1 \u003d \\ FRAC (4-25) (25) \\ lt 0 $, i.e. The second factor is a negative constant, and when dividing it, the sign of inequality will change:

\\ [\\ begin (align) & x + 1-0 \\ le 0 \\ rightarrow x \\ le -1; \\\\ & x \\ in \\ left (- \\ infty; -1 \\ right]. \\\\\\ End (Align) \\]

Finally, the last inequality from the current "kit":

\\ [((\\ left (\\ FRAC (27) (\\ SQRT (3)) \\ RIGHT)) ^ (- x)) \\ lt ((9) ^ (4-2x)) \\ CDOT 81 \\]

In principle, the idea of \u200b\u200bthe solution here is also clear: all the indicative functions included in the inequality must be reduced to the basis of "3". But for this will have to tinker with roots and degrees:

\\ [\\ begin (align) \\ FRAC (27) (\\ sqrt (3)) \u003d \\ FRAC (((3) ^ (3))) (((3) ^ (\\ FRAC (1) (3)) )) \u003d ((3) ^ (3- \\ FRAC (1) (3))) \u003d ((3) ^ (\\ FRAC (8) (3))); \\\\ & 9 \u003d ((3) ^ (2)); \\ quad 81 \u003d ((3) ^ (4)). \\\\\\ End (Align) \\]

Taking into account these facts, the initial inequality can be rewritten so:

\\ [\\ begin (align) & ((\\ left (((3) ^ (\\ FRAC (8) (3))) \\ Right)) ^ (- x)) \\ lt ((\\ left (((3) ^ (2)) \\ right)) ^ (4-2x)) \\ Cdot ((3) ^ (4)); \\\\ & ((3) ^ (- \\ FRAC (8X) (3))) \\ lt ((3) ^ (8-4x)) \\ Cdot ((3) ^ (4)); \\\\ & ((3) ^ (- \\ FRAC (8X) (3))) \\ lt ((3) ^ (8-4x + 4)); \\\\ & ((3) ^ (- \\ FRAC (8X) (3))) \\ lt ((3) ^ (4-4x)). \\\\\\ End (Align) \\]

Pay attention to the 2nd and 3rd line of calculations: before doing something with inequality, be sure to bring it to the same thing about which we spoke from the very beginning of the lesson: $ ((a) ^ (x)) \\ lt ((a) ^ (n)) $. As long as you have left or right, there are some left multipliers, additional constants, etc., no rationalization and "grinding" reason can not be performed! Countless tasks were improperly due to the misunderstanding of this simple fact. I myself am constantly watching this problem at my students, when we just proceed to the analysis of indicative and logarithmic inequalities.

But let's return to our task. Let's try this time to do without rationalization. We remember: the base of the degree is more than the unit, so the troika can simply shove - the sign of inequality will not change. We get:

\\ [\\ begin (Align) & - \\ FRAC (8x) (3) \\ lt 4-4x; \\\\ & 4x- \\ FRAC (8X) (3) \\ LT 4; \\\\ \\ FRAC (4X) (3) \\ LT 4; \\\\ & 4x \\ lt 12; \\\\ & X \\ LT 3. \\\\\\ End (Align) \\]

That's all. Final answer: $ x \\ in \\ left (- \\ infty; 3 \\ right) $.

Selecting a stable expression and replacement of variable

In conclusion, I propose to solve another four demonstration inequalities that are already quite complex for unprepared students. To cope with them, you need to remember the rules for working with degrees. In particular, the issuance of general factors for brackets.

But the most important thing is to learn to understand: what exactly can be taken out of brackets. Such an expression is called stable - it can be denoted by a new variable and thus get rid of the indicative function. So, let's look at the tasks:

\\ [\\ begin (align) & ((5) ^ (x + 2)) + ((5) ^ (x + 1)) \\ GE 6; \\\\ & ((3) ^ (x)) + ((3) ^ (x + 2)) \\ GE 90; \\\\ & ((25) ^ (x + 1.5)) - ((5) ^ (2x + 2)) \\ GT 2500; \\\\ \\ (\\ left (0.5 \\ right)) ^ (- 4x-8)) - ((16) ^ (x + 1.5)) \\ gt 768. \\\\\\ End (Align) \\]

Let's start with the very first line. We write down this inequality separately:

\\ [((5) ^ (x + 2)) + ((5) ^ (x + 1)) \\ GE 6 \\]

Note that $ ((5) ^ (x + 2)) \u003d (((5) ^ (x + 1 + 1)) \u003d (((5) ^ (x + 1)) \\ CDOT $ 5, so the right-hand part can rewrite:

Note that there are no other indicative functions except $ ((5) ^ (x + 1)) $, there is no inequality. And in general, nowhere else is there any variable $ x $, so we introduce a new variable: $ ((5) ^ (x + 1)) \u003d T $. We get the following design:

\\ [\\ begin (align) & 5t + t \\ ge 6; \\\\ & 6T \\ GE 6; \\\\ & T \\ GE 1. \\\\\\ End (Align) \\]

We return to the initial variable ($ T \u003d (((5) ^ (x + 1)) $), and at the same time we remember that 1 \u003d 5 0. We have:

\\ [\\ begin (align) & ((5) ^ (x + 1)) \\ GE ((5) ^ (0)); \\\\ & x + 1 \\ GE 0; \\\\ & X \\ GE -1. \\\\\\ End (Align) \\]

That's all the decision! Answer: $ x \\ in \\ left [-1; + \\ infty \\ right) $. Go to the second inequality:

\\ [((3) ^ (x)) + ((3) ^ (x + 2)) \\ GE 90 \\]

Here is all the same. Note that $ ((3) ^ (x + 2)) \u003d (((3) ^ (x)) \\ Cdot (((3) ^ (2)) \u003d 9 \\ Cdot ((3) ^ (x)) $ . Then the left part can be rewritten:

\\ [\\ Begin (Align) & ((3) ^ (x)) + 9 \\ Cdot ((3) ^ (x)) \\ GE 90; \\ quad \\ left | ((3) ^ (x)) \u003d T \\ Right. \\\\ & t + 9t \\ GE 90; \\\\ & 10t \\ GE 90; \\\\ & t \\ ge 9 \\ rightarrow ((3) ^ (x)) \\ GE 9 \\ rightarrow ((3) ^ (x)) \\ ge ((3) ^ (2)); \\\\ & X \\ GE 2 \\ Rightarrow X \\ IN \\ LEFT [2; + \\ infty \\ right). \\\\\\ End (Align) \\]

This is how it is necessary to make a decision on real control and independent work.

Well, let's try something more complicated. For example, here is inequality:

\\ [((25) ^ (x + 1.5)) - ((5) ^ (2x + 2)) \\ gt 2500 \\]

What is the problem here? First of all, the bases of the indicative functions standing on the left, different: 5 and 25. However, 25 \u003d 5 2, so the first term can be converted:

\\ [\\ begin (align) & ((25) ^ (x + 1.5)) \u003d ((\\ left ((((5) ^ (2)) \\ Right)) ^ (x + 1.5)) \u003d ((5) ^ (2x + 3)); \\\\ & ((5) ^ (2x + 3)) \u003d ((5) ^ (2x + 2 + 1)) \u003d ((5) ^ (2x + 2)) \\ CDOT 5. \\\\ End (align ) \\]

As you can see, first we all led to same baseAnd then noticed that the first term easily comes down to the second - just decompose the indicator. Now you can safely introduce a new variable: $ ((5) ^ (2x + 2)) \u003d T $, and all inequality will rewrite:

\\ [\\ begin (Align) & 5T-T \\ GE 2500; \\\\ & 4t \\ GE 2500; \\\\ & t \\ ge 625 \u003d ((5) ^ (4)); \\\\ & ((5) ^ (2x + 2)) \\ GE ((5) ^ (4)); \\\\ & 2x + 2 \\ GE 4; \\\\ & 2x \\ GE 2; \\\\ & X \\ GE 1. \\\\\\ End (Align) \\]

And again no difficulties! Final answer: $ x \\ in \\ left [1; + \\ infty \\ right) $. Go to the final inequality in today's lesson:

\\ [((\\ left (0.5 \\ right)) ^ (- 4x-8)) - ((16) ^ (x + 1.5)) \\ gt 768 \\]

The first thing to pay attention is, of course, decimal Based on the first degree. It is necessary to get rid of it, and at the same time bring all the indicative functions to the same base - the number "2":

\\ [\\ begin (align) & 0.5 \u003d \\ frac (1) (2) \u003d ((2) ^ (- 1)) \\ rightarrow ((\\ left (0.5 \\ right)) ^ (- 4x- 8)) \u003d ((\\ left (((2) ^ (- 1)) \\ Right)) ^ (- 4x-8)) \u003d ((2) ^ (4x + 8)); \\\\ & 16 \u003d ((2) ^ (4)) \\ rightarrow ((16) ^ (x + 1.5)) \u003d ((\\ left ((((2) ^ (4)) \\ Right)) ^ ( x + 1.5)) \u003d ((2) ^ (4x + 6)); \\\\ & ((2) ^ (4x + 8)) - ((2) ^ (4x + 6)) \\ gt 768. \\\\\\ End (Align) \\]

Excellent, we made the first step - everything led to the same basis. Now it is necessary to allocate a stable expression. Note that $ ((2) ^ (4x + 8)) \u003d ((2) ^ (4x + 6 + 2)) \u003d ((2) ^ (4x + 6)) \\ CDOT $ 4. If you enter a new variable $ ((2) ^ (4x + 6)) \u003d T $, then the initial inequality can be rewritten as follows:

\\ [\\ begin (align) & 4t-t \\ gt 768; \\\\ & 3t \\ gt 768; \\\\ & t \\ gt 256 \u003d ((2) ^ (8)); \\\\ & ((2) ^ (4x + 6)) \\ gt ((2) ^ (8)); \\\\ & 4x + 6 \\ GT 8; \\\\ & 4x \\ gt 2; \\\\ & X \\ GT \\ FRAC (1) (2) \u003d 0.5. \\\\\\ End (Align) \\]

Naturally, the question may arise: how did we find that 256 \u003d 2 8? Unfortunately, you just need to know the degree deductions (and at the same time the degree of triple and five). Well, or divide 256 to 2 (it is possible to divide, since a 256-point number) until we get the result. It will look like this:

\\ [\\ Begin (Align) & 256 \u003d 128 \\ CDOT 2 \u003d \\\\ & \u003d 64 \\ CDot 2 \\ CDot 2 \u003d \\\\ & \u003d 32 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \u003d \\\\ & \u003d 16 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \u003d \\\\ & \u003d 8 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \u003d \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ Cdot 2 \u003d \\\\ & \u003d 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \\ CDOT 2 \u003d \\\\ & \u003d ((2) ^ (8)). \\ ENG (Align ) \\]

The same with a triple (numbers 9, 27, 81 and 243 are its degrees), and with a seven (Numbers 49 and 343, too, would also be good to remember). Well, and the tops also have "beautiful" degrees to know:

\\ [\\ begin (align) & ((5) ^ (2)) \u003d 25; \\\\ & ((5) ^ (3)) \u003d 125; \\\\ & ((5) ^ (4)) \u003d 625; \\\\ & ((5) ^ (5)) \u003d 3125. \\\\\\ End (Align) \\]

Of course, all these numbers can be restored in the mind if desired, simply multiplying them to each other. However, when you have to solve several demonstration inequalities, and each next is more complicated than the previous one, then the latter, what I want to think - it is some numbers there. And in this sense, these tasks are more complex than the "classic" inequalities that are solved by the interval method.

In this lesson, we will look at various demonstration inequalities and learn them to decide, based on the method of solving the simplest demonstration inequalities

1. Definition and properties of the indicative function

Recall the definition and basic properties of the indicative function. It is on the properties that the solution of all indicatory equations and inequalities.

Exponential function - this is a function of the form where the basis of the degree and here x is an independent variable, the argument; y - dependent variable, function.

Fig. 1. schedule indicative function

The graph shows increasing and decreasing exhibitors illustrating an indicative function at the basis of a larger unit and a smaller unit, but a large zero, respectively.

Both curves pass through the point (0; 1)

Properties of the indicative function:

Domain: ;

Value area:;

The function of the monotonne, with increasing, with decreases.

The monotone feature takes each of its own value with the only value of the argument.

When the argument increases from minus to plus infinity, the function increases from zero not inclusive to plus infinity, i.e., with these argument values, we have a monotonously increasing function (). With the opposite, when the argument increases from minus to plus infinity, the function decreases from infinity to zero is not inclusive, i.e., with these values \u200b\u200bof the argument, we have a monotonously decreasing function ().

2. The simplest demonstration inequalities, decision technique, example

Based on the foregoing, we give the method of solving the simplest demonstration inequalities:

Methods of inequality solutions:

Equalize the bases of degrees;

Compare indicators, saving or changing to the opposite sign of inequality.

The solution of complex demonstration inequalities is, as a rule, in their information to the simplest indicative inequalities.

The foundation of the degree is greater than the unit, it means that the sign of inequality is maintained:

We transform the right side according to the degree properties:

The foundation of the degree is less than the unit, the sign of inequality must be changed to the opposite:

To solve square inequality, the corresponding square equation is solved:

On the Vieta Theorem find roots:

Parabola branches are directed up.

Thus, we have the solution of inequality:

It is easy to guess that the right part can be represented as a degree with a zero indicator:

The foundation of the degree is more united, the sign of inequality does not change, we get:

Recall the method of solving such inequalities.

We consider a fractional rational function:

Find the area of \u200b\u200bdefinition:

We find the roots of the function:

The function has the only root,

Select the intervals of the alignment and determine the signs of the function at each interval:

Fig. 2. Sign intervals

Thus, they received the answer.

Answer:

3. Solution of typical indicative inequalities

Consider inequalities with the same indicators, but various bases.

One of the properties of the indicative function - it takes strictly positive values \u200b\u200bwith any values \u200b\u200bof the argument, it means that an indicative function can be divided. Perform the division of a given inequality to the right part of it:

The foundation of the degree is more united, the sign of inequality is preserved.

We illustrate the solution:

Figure 6.3 shows graphs of functions and. Obviously, when the argument is greater than zero, the function graph is located above, this feature is greater. When the values \u200b\u200bof the argument are negative, the function passes below, it is less. The value of the function argument is equal, it means that this point is also a solution to the specified inequality.

Fig. 3. Illustration for example 4

We convert the predetermined inequality according to the degree properties:

We give similar members:

We split both parts on:

Now we continue to solve analogously to example 4, we divide both parts on:

The foundation of the degree is more united, the sign of inequality is maintained:

4. Graphic solution of indicative inequalities

Example 6 - Solve inequality graphically:

Consider the functions in the left and right part and build a schedule of each of them.

The function is an exhibitor, increases throughout its definition area, i.e., with all valid values \u200b\u200bof the argument.

The function is linear, decreases throughout its definition area, i.e., with all valid values \u200b\u200bof the argument.

If these functions intersect, that is, the system has a solution, then such a solution is the only one and it is easy to guess. To do this, go through integers ()

It is easy to see that the root of this system is:

Thus, graphs of functions intersect at a point with an argument equal to one.

Now you need to get the answer. The meaning of the specified inequality is that the exhibitor should be greater than or equal to the linear function, that is, be higher or coincided with it. Obvious answer: (Figure 6.4)

Fig. 4. Illustration for example 6

So, we reviewed the decision different typeshY indicative inequalities. Next, we proceed to the consideration of more complex demonstration inequalities.

Bibliography

Mordkovich A. G. Algebra and began mathematical analysis. - M.: Mnemozin. Muravin G. K., Muravina O. V. Algebra and began mathematical analysis. - M.: Drop. Kolmogorov A.N., Abramov A. M., Dudnitsyn Yu. P. and others. Algebra and began mathematical analysis. - M.: Enlightenment.

Math. MD. Mathematics-Repetition. COM. Diffur. kemsu. RU.

Homework

1. Algebra and start analysis, 10-11 class (A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn) 1990, No. 472, 473;

2. Solve inequality:

3. Solve inequality.