Đáp án:
\(\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 1} \right)\left( {5\sqrt x - 11} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
A = \left( {1 - \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right):\left( {\dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{\sqrt x - 2}}{{3 - \sqrt x }} + \dfrac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}} \right)\\
= \dfrac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x + 1}}:\left[ {\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - {{\left( {\sqrt x - 2} \right)}^2} + \sqrt x + 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}} \right]\\
= \dfrac{1}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{x - 9 - x + 4\sqrt x - 4 + \sqrt x + 2}}\\
= \dfrac{1}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{5\sqrt x - 11}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 1} \right)\left( {5\sqrt x - 11} \right)}}
\end{array}\)
