Đáp án:
\[\left[ \begin{array}{l}
x \ge 3\\
1 < x \le 2
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge \frac{1}{2}\\
x \ne 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{{\left( {\sqrt {x + 1} - \sqrt {2x - 1} } \right)\left( {\sqrt {x + 1} - 2} \right)}}{{x - 1}} \le 0\\
\Leftrightarrow \frac{{\frac{{\left( {x + 1} \right) - \left( {2x - 1} \right)}}{{\sqrt {x + 1} + \sqrt {2x - 1} }}.\frac{{x + 1 - 4}}{{\sqrt {x + 1} + 2}}}}{{x - 1}} \le 0\\
\Leftrightarrow \frac{{\left( {2 - x} \right).\left( {x - 3} \right)}}{{\left( {\sqrt {x + 1} + \sqrt {2x - 1} } \right).\left( {\sqrt {x + 1} + 2} \right)\left( {x - 1} \right)}} \le 0\,\,\,\,\,\left( 1 \right)\\
\sqrt {x + 1} + \sqrt {2x - 1} > 0,\,\,\,\,\forall x \ge \frac{1}{2}\\
\sqrt {x + 1} + 2 > 0,\,\,\,\,\forall x \ge \frac{1}{2}\\
\left( 1 \right) \Leftrightarrow \frac{{\left( {2 - x} \right)\left( {x - 3} \right)}}{{x - 1}} \le 0\\
\Leftrightarrow \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
1 < x \le 2
\end{array} \right.\left( {t/m} \right)
\end{array}\)
