Problem 8. Solve the inequality

$$
\frac{2 \sqrt{x+1}}{1-2 \sqrt{3-x}}<1
$$
Solution. The expression $2 \sqrt{x+1}$ is defined on the half-interval $[-1 ;+\infty)$. The expression $1-2 \sqrt{3-x}$ is defined and non-zero on the intervals ( $-\infty$; $2.75$ ) and ( 2, $75 ; 3]$, so the inequality (1) is defined on the intervals $[-1 ; 2.75$ ) and $(2.75 ; 3]$.

We construct the graphs of the functions: $f(x)=2 \sqrt{x+1}$ and $\varphi(x)=$ $=1-2 \sqrt{3-x}$ on the intervals $(-1 ; 2.75)$ and $(2.75 ; 3]$.

Since on the half-interval $[-1 ; 2.75)$ the function $f(x)$ is non-negative, then $\frac{f(x)}{\varphi(x)} \leqslant 0$ on this interval and the half-interval $[-1 ; 2.75)$ is a solution to the inequality ( 1 ).

The functions $f(x)$ and $\varphi(x)$ on the half-interval $(2.75 ; 3]$ are positive and increasing: $\sqrt[4]{15}$ $>\sqrt[4]{15}: 1>1$.

Answer. $[-1 ; 2.75)$.