Giải thích các bước giải:
\[\begin{array}{l}
DK:\left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\\
a,\\
Q = \frac{{\sqrt x + 2}}{{\sqrt x - 3}} - \frac{{\sqrt x + 1}}{{\sqrt x - 2}} - \frac{{3\left( {\sqrt x - 1} \right)}}{{x - 5\sqrt x + 6}}\\
\Leftrightarrow Q = \frac{{\sqrt x + 2}}{{\sqrt x - 3}} - \frac{{\sqrt x + 1}}{{\sqrt x - 2}} - \frac{{3\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
\Leftrightarrow Q = \frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right) - 3\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
\Leftrightarrow Q = \frac{{x - 4 - \left( {x - 2\sqrt x - 3} \right) - 3\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} = \frac{{ - \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} = \frac{1}{{3 - \sqrt x }}
\end{array}\]
b,\[\begin{array}{l}
Q < 1 \Leftrightarrow \frac{1}{{3 - \sqrt x }} - 1 < 0\\
\Leftrightarrow \frac{{1 - 3 + \sqrt x }}{{3 - \sqrt x }} < 0 \Leftrightarrow \frac{{\sqrt x - 2}}{{\sqrt x - 3}} > 0 \Leftrightarrow \left[ \begin{array}{l}
x > 9\\
x < 4
\end{array} \right.
\end{array}\]
