2. Solve the inequality $\frac{\sqrt{x}-2}{1-\sqrt{x+1}} \geq 1+\sqrt{x+1}$.
Answer: $x \in(0 ; 1]$.

Solution. The domain of the inequality is determined by the conditions $x \geq 0,1-\sqrt{x+1} \neq 0$, from which we get that $x>0$. 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

$$
\sqrt{x}-2 \leq 1-(x+1) \Leftrightarrow \sqrt{x} \leq 2-x \stackrel{\text { ○d3 }}{\Leftrightarrow}\left\{\begin{array} { l } 
{ 2 - x \geq 0 , } \\
{ x \leq 4 - 4 x + x ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x \leq 2 , } \\
{ x ^ { 2 } - 5 x + 4 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \leq 2, \\
x \in(-\infty ; 1] \cup[4 ;+\infty),
\end{array}\right.\right.\right.
$$

from which $x \leq 1$. Taking into account the domain, we finally get $x \in(0 ; 1]$.