Đáp án:
\(\begin{array}{l}
a)\,\,M = - \frac{1}{{\sqrt x - 3}}\\
b)\,\,9 < x < 16
\end{array}\)
Giải thích các bước giải:
a)
\(\begin{array}{l}M = \frac{{\sqrt x + 2}}{{\sqrt x - 3}} - \frac{{\sqrt x + 1}}{{\sqrt x - 2}} - 3\frac{{\sqrt x - 1}}{{x - 5\sqrt x + 6}}\,\,\left( {x \ge 0,\,\,x \ne 4,\,\,x \ne 9} \right)\\M = \frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right) - 3\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\M = \frac{{x - 4 - x + 3\sqrt x - \sqrt x + 3 - 3\sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\M = \frac{{ - \sqrt x + 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\M = \frac{{ - 1}}{{\sqrt x - 3}}\end{array}\)
b)
\(\begin{array}{l}M < - 1\\ \Leftrightarrow \frac{{ - 1}}{{\sqrt x - 3}} < - 1\\ \Leftrightarrow \frac{{ - 1}}{{\sqrt x - 3}} + 1 < 0\\ \Leftrightarrow \frac{{ - 1 + \sqrt x - 3}}{{\sqrt x - 3}} < 0\\ \Leftrightarrow \frac{{\sqrt x - 4}}{{\sqrt x - 3}} < 0\\ \Leftrightarrow 3 < \sqrt x < 4\\ \Leftrightarrow 9 < x < 16\end{array}\)
