Đáp án:
\(\dfrac{{x\sqrt x - 4x - 15\sqrt x + 8}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne \left\{ {1;4} \right\}\\
\dfrac{{\sqrt x - 4}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}} - \dfrac{5}{{\sqrt x - 2}} + \dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 4} \right)\left( {\sqrt x - 2} \right) - 5\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) + \sqrt x \left( {x - 4} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 6\sqrt x + 8 - 5\left( {x + \sqrt x - 2} \right) + x\sqrt x - 4\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x\sqrt x + x - 10\sqrt x + 8 - 5x - 5\sqrt x + 10}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x\sqrt x - 4x - 15\sqrt x + 8}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}
\end{array}\)
