Đáp án:
Giải thích các bước giải:
ĐK: $x\ge 0; x\ne 1$
Ta có:
$\begin{array}{l}
A = \left( {\dfrac{{2\sqrt x }}{{x\sqrt x + x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x + 1}}} \right):\left( {\dfrac{{2\sqrt x }}{{\sqrt x + 1}} - 1} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}} - \dfrac{1}{{\sqrt x + 1}}} \right):\left( {\dfrac{{2\sqrt x - \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right)\\
= \left( {\dfrac{{2\sqrt x - \left( {x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}} \right):\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{ - \left( {x - 2\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{ - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{ - \left( {\sqrt x - 1} \right)}}{{x + 1}}\\
= \dfrac{{1 - \sqrt x }}{{x + 1}}
\end{array}$
Vậy $A = \dfrac{{1 - \sqrt x }}{{x + 1}}$ khi $x\ge 0; x\ne 1$
