Đáp án: $\left[ \begin{array}{l}
x = - 1\\
\dfrac{3}{2} \le x \le 4
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
\sqrt { - {x^2} + 3x + 4} \le x + 1\left( 1 \right)\\
Dkxd:\left\{ \begin{array}{l}
- {x^2} + 3x + 4 \ge 0\\
x + 1 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} - 3x - 4 \le 0\\
x \ge - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x - 4} \right)\left( {x + 1} \right) \le 0\\
x \ge - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 \le x \le 4\\
x \ge - 1
\end{array} \right.\\
\Rightarrow - 1 \le x \le 4\\
\left( 1 \right) \Rightarrow - {x^2} + 3x + 4 \le {\left( {x + 1} \right)^2}\\
\Rightarrow - {x^2} + 3x + 4 \le {x^2} + 2x + 1\\
\Rightarrow 2{x^2} - x - 3 \ge 0\\
\Rightarrow \left( {2x - 3} \right)\left( {x + 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x \le - 1
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = - 1\\
\dfrac{3}{2} \le x \le 4
\end{array} \right.
\end{array}$
