Đáp án:
$A = \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}},x > 0;x \ne 4$
Giải thích các bước giải:
ĐKXĐ: $x > 0;x \ne 4$
Ta có:
$\begin{array}{l}
A = \left( {\dfrac{{5 + 3\sqrt x }}{{x - \sqrt x - 2}} + \dfrac{{4 + 2\sqrt x }}{{4 - x}}} \right).\dfrac{{x + \sqrt x }}{{\sqrt x }}\\
= \left( {\dfrac{{3\sqrt x + 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{2\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }}\\
= \left( {\dfrac{{3\sqrt x + 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{2}{{\sqrt x - 2}}} \right).\left( {\sqrt x + 1} \right)\\
= \dfrac{{3\sqrt x + 5 - 2\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}
\end{array}$
Vậy $A = \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}},x > 0;x \ne 4$
