Đáp án:
\(\dfrac{x}{{\sqrt x - 5}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne 9\\
\left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{2x}}{{9 - x}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{x - 3\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\
= \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right) - 2x}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left[ {\dfrac{{\sqrt x - 1 - 2\left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right]\\
= \dfrac{{x - 3\sqrt x - 2x}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x - 1 - 2\sqrt x + 6}}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - x - 3\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{5 - \sqrt x }}\\
= \dfrac{{ - \sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{5 - \sqrt x }}\\
= \dfrac{x}{{\sqrt x - 5}}
\end{array}\)
