1. Solve the inequality

$$
\frac{4+\sqrt{-3-x}}{3-|x+4|} \leqslant 1
$$

In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and do not exceed 12 in absolute value.
Solution. Transferring 1 to the left side and bringing to a common denominator:

$$
\frac{1+\sqrt{-3-x}+|x+4|}{3-|x+4|} \leqslant 0
$$

For $x>-3$, the expression under the square root is negative, meaning there are no solutions. For $x \leqslant-3$, the numerator is positive, and thus,

$$
3-|x+4|3 \Longleftrightarrow x>-1
$$

or $x<-7$.

Therefore, the solution to the inequality is: $x<-7$. The required integer roots according to the condition are: $-12, -11, \ldots$, -8. Their sum is -50.

Answer: -50.