Đáp án:
\[\frac{{3\sqrt x }}{{\sqrt x + 3}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \frac{{2\sqrt x }}{{\sqrt x + 3}} + \frac{{\sqrt x + 1}}{{\sqrt x - 3}} + \frac{{7\sqrt x + 3}}{{9 - x}}\\
= \frac{{2\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} + \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} + \frac{{7\sqrt x + 3}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}\\
= \frac{{2x - 6\sqrt x + x + 4\sqrt x + 3 - 7\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{3x - 9\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{3\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} = \frac{{3\sqrt x }}{{\sqrt x + 3}}
\end{array}\)
