Đáp án:
\[P = \frac{2}{{a - 1}}\,\,\,\,\,\left( {a > 0,\,\,a \ne 1} \right)\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
P = \left( {\frac{{\sqrt a + 2}}{{a + 2\sqrt a + 1}} - \frac{{\sqrt a - 2}}{{a - 1}}} \right).\frac{{\sqrt a + 1}}{{\sqrt a }}\\
= \left( {\frac{{\sqrt a + 2}}{{{{\left( {\sqrt a + 1} \right)}^2}}} - \frac{{\sqrt a - 2}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}} \right).\frac{{\sqrt a + 1}}{{\sqrt a }}\\
= \left( {\frac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a - 2} \right)\left( {\sqrt a + 1} \right)}}{{{{\left( {\sqrt a + 1} \right)}^2}\left( {\sqrt a - 1} \right)}}} \right).\frac{{\sqrt a + 1}}{{\sqrt a }}\\
= \left( {\frac{{\left( {a + \sqrt a - 2} \right) - \left( {a - \sqrt a - 2} \right)}}{{{{\left( {\sqrt a + 1} \right)}^2}\left( {\sqrt a - 1} \right)}}} \right).\frac{{\sqrt a + 1}}{{\sqrt a }}\\
= \frac{{2\sqrt a }}{{{{\left( {\sqrt a + 1} \right)}^2}\left( {\sqrt a - 1} \right)}}.\frac{{\sqrt a + 1}}{{\sqrt a }}\\
= \frac{2}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}\\
= \frac{2}{{a - 1}}
\end{array}\)
Vậy \(P = \frac{2}{{a - 1}}\,\,\,\,\,\left( {a > 0,\,\,a \ne 1} \right)\)
