2. Solve the inequality $\frac{\sqrt{x}-6}{2-\sqrt{x+4}} \geq 2+\sqrt{x+4}$.
Answer: $x \in(0 ; 4]$.

Solution. The domain of the inequality is determined by the conditions $x \geq 0, 2-\sqrt{x+4} \neq 0$, 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

$$
\sqrt{x}-6 \leq 4-(x+4) \Leftrightarrow \sqrt{x} \leq 6-x \stackrel{\text {ODZ}}{\Leftrightarrow}\left\{\begin{array} { l } 
{ 6 - x \geq 0 , } \\
{ x \leq 3 6 - 1 2 x + x ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x \leq 6 , } \\
{ x ^ { 2 } - 1 3 x + 3 6 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \leq 6, \\
x \in(-\infty ; 4] \cup[9 ;+\infty),
\end{array}\right.\right.\right.
$$

from which $x \leq 4$. Taking into account the domain, we finally get $x \in(0 ; 4]$.