Đáp án:
\(\dfrac{3}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) + 2\sqrt x \left( {\sqrt x + 3} \right) - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \dfrac{{3\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \dfrac{3}{{\sqrt x + 3}}
\end{array}\)
