Example 18. Find all values of $a \neq 0$, for which the inequality

$$
a^{2}\left|a+x / a^{2}\right|+|1+x| \leqslant 1-a^{3}
$$

has at least four different solutions that are integers.
Solution. The left side of inequality (17) is non-negative for any values of $a$ and $x$; therefore, inequality

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-140.jpg?height=579&width=660&top_left_y=1361&top_left_x=1049)

Fig. 3.1 can have a solution only when its right side is non-negative, i.e., when $a^{3} \leq 1$. Since

$$
a^{2}\left|a+x / a^{2}\right|=\left|a^{3}+x\right|
$$

then inequality (17) is equivalent to the inequality

$$
\left|x+a^{3}\right|+|x+1| \leqslant 1-a^{3}
$$

Inequality (18) can be solved formally by expanding the absolute value signs. We will provide another solution. For this, rewrite it in the form

$$
\left|x-\left(-a^{3}\right)\right|+|x-(-1)| \leqslant 1-a^{3}
$$

According to the geometric interpretation of the absolute value, solving inequality (19) means finding all points $x$ on the coordinate line such that the sum of the distances from each of them to the point with coordinate $-a^{3}$ and to the point with coordinate -1 is not greater than $1-a^{3}$, i.e., not greater than the length of the segment $\left[-1 ;-a^{3}\right]$. The specified condition is satisfied by those and only those points for which the double inequality $-1 \leqslant x \leqslant-a^{3}$ holds. Thus, the solution to inequality (17) are all numbers from the segment $\left[-1 ;-a^{3}\right]$. For the segment $\left[-1 ;-a^{3}\right]$ to contain at least four integers, it must contain the numbers $-1,0,2,1$, i.e., the inequality $2 \leqslant-a^{3}$ must hold. From this inequality, it follows that only for $a \in$ $\in(-\infty ;-\sqrt{2}]$ does inequality (17) have at least four different solutions that are integers.

## TASK 1

Solve the inequality:

1) $0.3|x|-1 \leqslant \frac{6-|x|}{2}$; 2) $(|x|-3)(|x|+7)0$
6) $x^{2}-4 x-2|x-2|+1 \leqslant 0$;
7) $x^{2}+6 x-4|x+3|-12>0$;
8) $x^{2}-8 x-\frac{3}{|x-4|}+18 \leqslant 0$;
9) $x^{2}+10 x-\frac{5}{|x+5|}+4>0$;
10) $\left|\frac{x+4}{x+2}\right| \leqslant 1 ; \quad$ 12) $\left|\frac{x-3}{x-5}\right| \geqslant 1$.

## TASK 2

Solve the inequality:

1) $|3-x|x+4$
4) $3|x-1| \leqslant x+3$;
5) $x^{2}-7 x+12|x-6|$
7) $\left|x^{2}-4 x\right|4$
9) $\left|x^{2}+3 x\right| \geqslant 2-x^{2}$
10) $\left|x^{2}-6 x+8\right|1 / 2$;
2) $|x-2|3 x-x^{2}-2 ;$ 10) $\left|x^{2}-1\right|1$;
2) $\frac{|x+2|-x}{x}-1$;
4) $\frac{x^{2}-5 x+6}{|x|+7}2$
7) $\left|\frac{x^{2}-3 x-1}{x^{2}+x+1}\right|1$;
7) $\left|\frac{x^{2}-1}{x+2}\right|2+x-|3-x|$
4) $|x| \geqslant \frac{2 x}{|x-3|}$
5) $\frac{\left|x^{2}-4 x\right|+3}{x^{2}+|x-5|} \geqslant 1$
6) $|7-2 x||x-1| ; \quad$ 2) $2|x| \leqslant 4+|x+1|$;
2) $|2 x+3|>|x|-4 x-1 ; \quad$ 4) $\frac{4 x-1}{|x-1|} \geqslant|x+1|$;
3) $\left.\frac{\left|x^{2}-2 x\right|+4}{x^{2}+|x+2|} \geqslant 1 ; \quad 6\right)|5-x|1-x$
3) $\left|\frac{3|x|+2}{|x|-1}\right|1$.
2. For all $a$ solve the inequality:
1) $|x-a|-2 a>|x-3 a| ; \quad 2)|x+2 a|+|x-a|-1$;
2) $\left|x^{2}+21 x+34\right|5 ; \quad 5)|2 x-3| \leqslant 4$;
5) $|5 x-4| \geqslant 6$
6) $x^{2}-5|x|+61$
10) $\left|x^{2}-5 x\right| \leq 6$
11) $\left|-4 x^{2}-6 x-5\right| \geqslant 9$;
12) $\left|x^{2}-2 x\right|9 x-2$;
15) $x^{2}+4 \geqslant|3 x+2|-7 x$
16) $x^{2}-|5 x+8|>0$
17) $3|x-1|+x^{2}-7>0$;
18) $|x-6| \geqslant x^{2}-5 x+9$;
19) $\left|x^{2}-x-8\right| \leqslant x$
20) $\left|x^{2}-3\right|+2 x+1 \geqslant 0$
21) $(1+x)^{2} \leqslant\left|1-x^{2}\right|$
22) $\left|x^{2}-2 x-3\right|x+3$
24) $\left|x^{2}-6 x+8\right| \leqslant 4-x$
25) $\left|x^{2}-1\right| \geqslant x^{2}+1$
26) $\left|\frac{2 x-1}{x+2}\right| \leqslant 4$
27) $\left|\frac{x^{2}-2 x+1}{x-3}\right|>1$.
2. Solve the inequality:
1) $|2 x-1||1+x / 5|$
5) $|x-1|-|x|+|2 x+3|>2 x+4$;
6) $\left|\frac{-5}{x+2}\right|4$;
8) $|x+2|+|x+1|+|x-4| \geqslant 9$
9) $|x-1|-2|x-2|+3|x-3| \leqslant 4$
10) $|x-2|+|x-3|+|2 x-8||x+2|+|x|-3$;
13) $|x-1|-|x-2|+|x-3|2 x^{2}-x$; 19) $\left|x^{2}+6 x+8\right| \leqslant-x^{2}-6 x-8$;
17) $\left.\left|x^{2}+x-20\right| \leqslant x^{2}+x-20 ; \quad 21\right)|x-3|>\left|x^{2}-3\right|$;
18) $\left|2 x^{2}-x-10\right|>\left|x^{2}-8 x-22\right|$;
19) $\left|x^{2}-5\right| x|+4| \geqslant\left|2 x^{2}-3\right| x|+1| ;$
20) $\frac{(1+x)(2+x)}{x^{2}-|x|-2} \geqslant-3 x$
21) $\frac{|2 x-1|}{x^{2}+x-2} \geq 3$
22) ||$x^{2}-3 x+2|-1|>x-2$;
23) $\frac{\left|x^{2}-2 x\right|-1-2 x}{