Đáp án:
\(4 \le x \le 9\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge - \dfrac{7}{2}\\
x - \sqrt {2x + 7} \le 4\\
\to x - 4 \le \sqrt {2x + 7} \\
\to \left\{ \begin{array}{l}
x \ge 4\\
{x^2} - 8x + 16 \le 2x + 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 4\\
{x^2} - 10x + 9 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 4\\
\left( {x - 9} \right)\left( {x - 1} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 4\\
x \in \left[ {1;9} \right]
\end{array} \right.\\
KL:4 \le x \le 9
\end{array}\)
