1. Solve the inequality

$$
\frac{2^{2+\sqrt{x-1}}-24}{2^{1+\sqrt{x-1}}-8}>1 .
$$

In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-70 ; 34)$.
Solution. Let $t=2^{1+\sqrt{x-1}}$, we get

$$
\frac{2 t-24}{t-8} \geqslant 1 \Longleftrightarrow \frac{t-16}{t-8} \geqslant 0 \Longleftrightarrow t \in(-\infty ; 8) \cup[16 ;+\infty)
$$

From this, either $2^{1+\sqrt{x-1}}<2^{3}, \sqrt{x-1}<2, x \in[1 ; 5)$, or $2^{1+\sqrt{x-1}} \geqslant 2^{4}, \sqrt{x-1} \geqslant 3, x \in[10 ;+\infty)$. Therefore, the solution to the inequality is $x \in[1 ; 5) \cup(10 ;+\infty)$. The desired sum is

$$
1+2+3+4+(10+11+12+\ldots+33)=10+\frac{10+33}{2} \cdot 24=526
$$

Answer: 526.