4. Solve the inequality

$$
\frac{x(x-2)(x-3)-(x-2)^{2}+1}{(|x-1|-|x-2|) \sqrt{16-x^{2}}} \geq 0
$$

points)
Solution. Domain of definition: $\left\{\begin{array}{l}|x-1| \neq|x-2|, \\ 16-x^{2}>0,\end{array} \Rightarrow\left\{\begin{array}{l}x \neq 3 / 2, \\ x \in(-4,4) .\end{array}\right.\right.$

Transform the numerator

$x(x-2)(x-3)+1-(x-2)^{2}=x(x-2)(x-3)+(1-x+2)(1+x-2)=$

$=x(x-2)(x-3)+(3-x)(x-1)=(x-3)\left(x^{2}-2 x+1-x\right)=(x-3)\left(x^{2}-3 x+1\right)$

Multiply the denominator by the strictly positive sum of absolute values

$$
\begin{aligned}
& \frac{(x-3)\left(x^{2}-3 x+1\right)}{(|x-1|-|x-2|)(|x-1|+|x-2|) \sqrt{16-x^{2}}} \geq 0, \\
& \frac{(x-3)\left(x^{2}-3 x+1\right)}{\left(|x-1|^{2}-|x-2|^{2}\right) \sqrt{16-x^{2}}} \geq 0, \frac{(x-3)\left(x^{2}-3 x+1\right)}{(x-1-x+2)(x-1+x-2) \sqrt{16-x^{2}}} \geq 0, \\
& \frac{(x-3)(x-3 / 2-\sqrt{5} / 2)(x-3 / 2+\sqrt{5} / 2)}{(2 x-3) \sqrt{16-x^{2}}} \geq 0
\end{aligned}
$$

Solving by the method of intervals, we get the answer: $x \in\left(-4 ; \frac{3-\sqrt{5}}{2}\right] \cup\left(\frac{3}{2} ; \frac{3+\sqrt{5}}{2}\right] \cup[3 ; 4)$.