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