Đáp án:
\[\left[ \begin{array}{l}
x < \frac{1}{2}\\
x \ge 2
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ:
\(\left\{ \begin{array}{l}
x + 2 \ge 0\\
3 - x \ge 0\\
5 - 2x \ge 0\\
x + 2 \ne 3 - x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 2\\
x \le 3\\
x \le \frac{5}{2}\\
x \ne \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 2 \le x \le \frac{5}{2}\\
x \ne \frac{1}{2}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{1}{{\sqrt {x + 2} - \sqrt {3 - x} }} \le \frac{1}{{\sqrt {5 - 2x} }}\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 2} - \sqrt {3 - x} < 0\\
\left\{ \begin{array}{l}
\sqrt {x + 2} - \sqrt {3 - x} > 0\\
\sqrt {x + 2} - \sqrt {3 - x} \ge \sqrt {5 - 2x}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 < 3 - x\\
\left\{ \begin{array}{l}
x + 2 > 3 - x\\
\sqrt {x + 2} \ge \sqrt {3 - x} + \sqrt {5 - 2x}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{2}\\
\left\{ \begin{array}{l}
x > \frac{1}{2}\\
x + 2 \ge 3 - x + 2\sqrt {\left( {3 - x} \right)\left( {5 - 2x} \right)} + 5 - 2x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{2}\\
\left\{ \begin{array}{l}
x > \frac{1}{2}\\
2x - 3 \ge \sqrt {\left( {3 - x} \right)\left( {5 - 2x} \right)}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{2}\\
\left\{ \begin{array}{l}
x > \frac{1}{2}\\
x \ge \frac{3}{2}\\
4{x^2} - 12x + 9 \ge 2{x^2} - 11x + 15
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{2}\\
\left\{ \begin{array}{l}
x \ge \frac{3}{2}\\
2{x^2} - x - 6 \ge 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{2}\\
\left\{ \begin{array}{l}
x \ge \frac{3}{2}\\
\left[ \begin{array}{l}
x \ge 2\\
x \le - \frac{3}{2}
\end{array} \right.
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x < \frac{1}{2}\\
x \ge 2
\end{array} \right.
\end{array}\)
