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