5. Solve the inequality: $\frac{\left(\left|x^{2}-2\right|-7\right)(|x+3|-5)}{|x-3|-|x-1|}>0$. Write the largest integer that is a solution to the inequality in your answer.
5. $\frac{\left(\left|x^{2}-2\right|-7\right)\left(\left|x^{2}-2\right|+7\right)(|x+3|-5)(|x+3|+5)}{(|x-3|-|x-1|)(|x-3|+|x-1|)}>0 \Leftrightarrow \frac{\left(\left(x^{2}-2\right)^{2}-49\right)\left((x+3)^{2}-25\right)}{(x-3)^{2}-(x-1)^{2}}>0$ $\Leftrightarrow \frac{\left(x^{2}-2-7\right)\left(x^{2}-2+7\right)(x+3-5)(x+3+5)}{(x-3-x+1)(x-3+x-1)}>0 \Leftrightarrow \frac{(x-3)(x+3)(x-2)(x+8)}{(-2)(2 x-4)}>0$, $x \neq 2 \Leftrightarrow(x-3)(x+3)(x+8)<0$

, Solve using the interval method:

![](https://cdn.mathpix.com/cropped/2024_05_06_a53e162167c41a98f5ecg-1.jpg?height=155&width=874&top_left_y=2041&top_left_x=434)

From here, the solution is: $x<-8 ;-3<x<3 ; x \neq 2$. Therefore, the answer is: $\{1\}$