Đáp án:
\(\dfrac{{\left( {x\sqrt x + x - \sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\left( {x\sqrt x + 1} \right)\left( {x - 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 1\\
P = \left[ {\dfrac{{\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right].\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + x\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right].\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{x - \sqrt x + x\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{x\sqrt x + x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {x\sqrt x + x - \sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\left( {x\sqrt x + 1} \right)\left( {x - 1} \right)}}
\end{array}\)
