# 5. Solve the inequality

$$
\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left|x^{2}-4 x+2\right|-|x-2|\right) \leq 0
$$
# Solution:

$\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left|x^{2}-4 x+2\right|-|x-2|\right) \leq 0$. odZ: $2 x^{2}+7 x+3 \geq 0, \Rightarrow$

$x \in(-\infty ;-3] \cup[-0.5 ;+\infty)$. The original inequality is equivalent to the following

$\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left(x^{2}-4 x+2\right)^{2}-(x-2)^{2}\right) \leq 0 \Leftrightarrow$

$\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(x^{2}-5 x+4\right)\left(x^{2}-3 x\right) \leq 0 \Leftrightarrow$

$\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)(x-1)(x-4) x(x-3) \leq 0$.

If $x \leq -3$, then $3 x+4<0$, and $\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)<0$. Thus, we come to the inequality $(x-1)(x-4) x(x-3) \geq 0$, and $x \in(-\infty ;-3]$. If $x \geq -0.5$, then we come to the inequality $(\sqrt{2 x+1}-\sqrt{x+3})^{2}(x-1)(x-4) x(x-3) \leq 0$, and $x \in[0 ; 1] \cup\{2\} \cup[3 ; 4]$.

Finally, we have: Answer: $x \in(-\infty ;-3] \cup[0 ; 1] \cup\{2\} \cup[3 ; 4]$.