4. Solve the inequality $\left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left|x^{2}-4 x+2\right|-|x-2|\right) \leq 0$. (20 points)

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# Solution:

$$
\text { Domain: } 2 x^{2}+7 x+3 \geq 0, \Rightarrow x \in(-\infty ;-3] \cup[-0.5 ;+\infty) .
$$

The original inequality is equivalent to the following

$$
\begin{aligned}
& \left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(\left(x^{2}-4 x+2\right)^{2}-(x-2)^{2}\right) \leq 0 \Leftrightarrow \\
& \left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)\left(x^{2}-5 x+4\right)\left(x^{2}-3 x\right) \leq 0 \Leftrightarrow \\
& \left(3 x+4-2 \sqrt{2 x^{2}+7 x+3}\right)(x-1)(x-4) x(x-3) \leq 0
\end{aligned}
$$

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 arrive at the inequality $(x-1)(x-4) x(x-3) \geq 0$, and $x \in(-\infty ;-3]$.

If $x \geq -0.5$, then we arrive at the inequality

$$
(\sqrt{2 x+1}-\sqrt{x+3})^{2}(x-1)(x-4) x(x-3) \leq 0 \text {, and } x \in[0 ; 1] \cup\{2\} \cup[3 ; 4]
$$

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

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