Đáp án:
\(\dfrac{{\sqrt x - 6}}{{x - 4}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
Q = \dfrac{{x + 2\sqrt x - 10}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {x + 2\sqrt x - 10} \right)\left( {\sqrt x - 2} \right) - {{\left( {\sqrt x - 2} \right)}^2}\left( {\sqrt x + 2} \right) - x + \sqrt x + 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x\sqrt x - 2x + 2x - 4\sqrt x - 10\sqrt x + 20 - \left( {x - 4\sqrt x + 4} \right)\left( {\sqrt x + 2} \right) - x + \sqrt x + 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x\sqrt x - x - 13\sqrt x + 26 - x\sqrt x - 2x + 4x + 8\sqrt x - 4\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 9\sqrt x + 18}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 6} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 6}}{{x - 4}}
\end{array}\)
