`A =` ($\frac{2\sqrt{x} + x}{x\sqrt{x} -1}$ - $\frac{1}{\sqrt{x} - 1}$) : ($\frac{x - 1}{x + \sqrt{x} + 1}$) Rút gọn biểu thức `A`
1 câu trả lời
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{2\sqrt x + x}}{{x\sqrt x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 1}}{{x + \sqrt x + 1}}\\
= \left[ {\dfrac{{2\sqrt x + x}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x - 1}}} \right].\dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x + x - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{1}{{1 - x}}
\end{array}$