Đáp án:
\(\dfrac{{5\sqrt x }}{{2\sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \left( {\dfrac{2}{{\sqrt x - 2}} + \dfrac{3}{{2\sqrt x + 1}} - \dfrac{{5\sqrt x - 7}}{{2x - 3\sqrt x - 2}}} \right):\dfrac{{2\sqrt x + 3}}{{5x - 10\sqrt x }}\\
= \left[ {\dfrac{{4\sqrt x + 2 + 3\sqrt x - 6 - 5\sqrt x + 7}}{{\left( {\sqrt x - 2} \right)\left( {2\sqrt x + 1} \right)}}} \right].\dfrac{{5\sqrt x \left( {\sqrt x - 2} \right)}}{{2\sqrt x + 3}}\\
= \dfrac{{2\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {2\sqrt x + 1} \right)}}.\dfrac{{5\sqrt x \left( {\sqrt x - 2} \right)}}{{2\sqrt x + 3}}\\
= \dfrac{{5\sqrt x }}{{2\sqrt x + 1}}
\end{array}\)
