Đáp án:
\(\dfrac{{a + \sqrt a + 1}}{{\sqrt a - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:a \ge 0;a \ne 1\\
P = \left( {\dfrac{{a + 1 + \sqrt a }}{{a + 1}}} \right):\left[ {\dfrac{1}{{\sqrt a - 1}} - \dfrac{{2\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}} \right]\\
= \left( {\dfrac{{a + 1 + \sqrt a }}{{a + 1}}} \right):\dfrac{{a + 1 - 2\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\dfrac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}.\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + 1} \right)}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \dfrac{{a + \sqrt a + 1}}{{\sqrt a - 1}}
\end{array}\)
