5. Solve the inequality $\frac{|x-2|-\left|x^{2}-4 x+2\right|}{2 \sqrt{2 x^{2}+7 x+3}-3 x-4} \geq 0$.
# Solution:

$\frac{|x-2|-\left|x^{2}-4 x+2\right|}{2 \sqrt{2 x^{2}+7 x+3}-3 x-4} \geq 0 . \quad$ 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 $\frac{(x-2)^{2}-\left(x^{2}-4 x+2\right)^{2}}{2 \sqrt{2 x^{2}+7 x+3}-3 x-4} \geq 0 \Leftrightarrow \frac{\left(x^{2}-5 x+4\right)\left(x^{2}-3 x\right)}{3 x+4-2 \sqrt{2 x^{2}+7 x+3}} \geq 0 \Leftrightarrow \frac{(x-1)(x-4) x(x-3)}{3 x+4-2 \sqrt{2 x^{2}+7 x+3}} \geq 0$.

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 arrive at the inequality $(x-1)(x-4) x(x-3) \leq 0$, and, considering the condition $x \leq -3$, there are no solutions.

If $x \geq -0.5$, then we arrive at the inequality $\frac{(x-1)(x-4) x(x-3)}{(\sqrt{2 x+1}-\sqrt{x+3})^{2}} \geq 0$, and $x \in[-0.5 ; 0] \cup[1 ; 2) \cup(2 ; 3] \cup[4 ;+\infty)$.

Answer: $x \in[-0.5 ; 0] \cup[1 ; 2) \cup(2 ; 3] \cup[4 ;+\infty)$.