Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
Q = \dfrac{{x + 2\sqrt x - 10}}{{x - \sqrt x - 6}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} - \dfrac{1}{{\sqrt x - 2}}\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right)\\
= \dfrac{{x + 2\sqrt x - 10}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {x + 2\sqrt x - 10} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {x + 2\sqrt x - 10} \right) - \left( {x - 4} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{2\sqrt x - 6}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{2.\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{2}{{\sqrt x + 2}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{2.\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 6}}{{x - 4}}
\end{array}\)
