2. Solve the inequality $\frac{\sqrt{1-x}-12}{1-\sqrt{2-x}} \geq 1+\sqrt{2-x}$.
Answer: $x \in[-8 ; 1)$.

Solution. The domain of the inequality is determined by the conditions $x \leq 1,1-\sqrt{2-x} \neq 0$, from which we get that $x<1$. 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 sign of the inequality. Then

$$
\sqrt{1-x}-12 \leq 1-(2-x) \Leftrightarrow \sqrt{1-x} \leq 11+x \Leftrightarrow\left\{\begin{array} { l } 
{ 1 1 + x \geq 0 , } \\
{ 1 - x \leq 1 2 1 + 2 2 x + x ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x \geq - 1 1 , } \\
{ x ^ { 2 } + 2 3 x + 1 2 0 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \geq-11, \\
x \in(-\infty ;-15] \cup[-8 ;+\infty),
\end{array}\right.\right.\right.
$$

from which $x \geq-8$. Taking into account the domain, we finally get $x \in[-8 ; 1)$.