a) $Q= \left ( \dfrac{\sqrt{x} - 1}{\sqrt{x} +1} - \dfrac{\sqrt{x} + 1}{\sqrt{x} -1} \right ) \left ( \dfrac{1}{2 \sqrt{x}} - \dfrac{\sqrt{x}}{2}\right ) ^2 \\
= \dfrac{\left ( \sqrt{x} - 1 \right ) ^2 - \left ( \sqrt{x} + 1 \right ) ^2}{x-1} \left ( \dfrac{1-x}{2 \sqrt{x}}\right ) ^2 \\
= \dfrac{\left ( \sqrt{x} - 1 - \sqrt{x} - 1 \right ) \left ( \sqrt{x} - 1 + \sqrt{x} + 1 \right ) }{x-1} . \dfrac{(x-1)^2}{\left ( 2 \sqrt{x}\right ) ^2 } \\
= \dfrac{-4 \sqrt{x}}{x-1}. \dfrac{(x-1)^2}{4x} \\
= - \dfrac{x-1}{\sqrt{x}}$
b) $\dfrac{Q}{\sqrt{x}} > 2 \Leftrightarrow \left ( - \dfrac{x-1}{\sqrt{x}} \right ) . \dfrac{1}{\sqrt{x}} > 2$
$\Leftrightarrow - \dfrac{x-1}{x} >2 \\
\Leftrightarrow \dfrac{x-1}{x} < -2 \\
\Leftrightarrow 1 - \dfrac{1}{x} < -2 \\
\Leftrightarrow - \dfrac{1}{x} < -3 \\
\Leftrightarrow \dfrac{1}{x} > 3 \\
\Leftrightarrow x < \dfrac{1}{3}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
