Đáp án:
\[P = \dfrac{3}{{\sqrt a + 2}}\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
a \ge 0\\
a \ne 16
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
P = \dfrac{{\sqrt a .\left( {2\sqrt a + 1} \right)}}{{8 + 2\sqrt a - a}} + \dfrac{{\sqrt a + 4}}{{\sqrt a + 2}} - \dfrac{{\sqrt a + 2}}{{4 - \sqrt a }}\\
= \dfrac{{\sqrt a .\left( {2\sqrt a + 1} \right)}}{{\left( {8 + 4\sqrt a } \right) - \left( {2\sqrt a + a} \right)}} + \dfrac{{\sqrt a + 4}}{{\sqrt a + 2}} - \dfrac{{\sqrt a + 2}}{{4 - \sqrt a }}\\
= \dfrac{{2a + \sqrt a }}{{4.\left( {2 + \sqrt a } \right) - \sqrt a \left( {2 + \sqrt a } \right)}} + \dfrac{{\sqrt a + 4}}{{\sqrt a + 2}} - \dfrac{{\sqrt a + 2}}{{4 - \sqrt a }}\\
= \dfrac{{2a + \sqrt a }}{{\left( {2 + \sqrt a } \right)\left( {4 - \sqrt a } \right)}} + \dfrac{{\sqrt a + 4}}{{\sqrt a + 2}} - \dfrac{{\sqrt a + 2}}{{4 - \sqrt a }}\\
= \dfrac{{2a + \sqrt a + \left( {\sqrt a + 4} \right)\left( {4 - \sqrt a } \right) - {{\left( {\sqrt a + 2} \right)}^2}}}{{\left( {\sqrt a + 2} \right)\left( {4 - \sqrt a } \right)}}\\
= \dfrac{{2a + \sqrt a + 16 - a - \left( {a + 4\sqrt a + 4} \right)}}{{\left( {\sqrt a + 2} \right)\left( {4 - \sqrt a } \right)}}\\
= \dfrac{{ - 3\sqrt a + 12}}{{\left( {\sqrt a + 2} \right)\left( {4 - \sqrt a } \right)}}\\
= \dfrac{{3.\left( {4 - \sqrt a } \right)}}{{\left( {\sqrt a + 2} \right)\left( {4 - \sqrt a } \right)}}\\
= \dfrac{3}{{\sqrt a + 2}}
\end{array}\)
