2. Solve the inequality $\frac{\sqrt{2-x}-2}{1-\sqrt{3-x}} \geq 1+\sqrt{3-x}$.
Answer: $x \in[1 ; 2)$.

Solution. The domain of the inequality is determined by the conditions $x \leq 2,1-\sqrt{3-x} \neq 0$, from which we get that $x<2$. Note that on the domain, the denominator of the fraction is negative, so we can multiply both sides of the inequality by it, changing the inequality sign. Then

$$
\sqrt{2-x}-2 \leq 1-(3-x) \Leftrightarrow \sqrt{2-x} \leq x \Leftrightarrow\left\{\begin{array} { l } 
{ x \geq 0 , } \\
{ 2 - x \leq x ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x \geq 0 , } \\
{ x ^ { 2 } + x - 2 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \geq 0, \\
x \in(-\infty ;-2] \cup[1 ;+\infty),
\end{array}\right.\right.\right.
$$

from which $x \geq 1$. Taking into account the domain, we finally get $x \in[1 ; 2)$.