Đáp án: $Q = 2$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 1\\
Q = \left( {\dfrac{{5\sqrt x }}{{x - 1}} + \dfrac{3}{{2\sqrt x + 2}} - \dfrac{5}{{2\sqrt x - 2}}} \right):\dfrac{2}{{\sqrt x + 1}}\\
= \dfrac{{5\sqrt x .2 + 3\left( {\sqrt x - 1} \right) - 5\left( {\sqrt x + 1} \right)}}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{10\sqrt x + 3\sqrt x - 3 - 5\sqrt x - 5}}{{4\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{8\sqrt x - 8}}{{4\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{8\left( {\sqrt x - 1} \right)}}{{4\left( {\sqrt x - 1} \right)}}\\
= 2
\end{array}$
