Đáp án:
\(\dfrac{{\sqrt x + 2}}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{x - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 1}}{{\sqrt x \left( {x + \sqrt x + 1} \right)}} + \dfrac{{2x - 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {x - 2\sqrt x } \right)\sqrt x + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 2x - 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - 2x + x - 1 + 2x - 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x + x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{x + \sqrt x + 1}}
\end{array}\)
