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