Đáp án:
\[P = \frac{1}{{a - \sqrt {ab} + b}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \left( {\frac{1}{{\sqrt a + \sqrt b }} + \frac{{3\sqrt {ab} }}{{a\sqrt a + b\sqrt b }}} \right)\left( {\frac{1}{{\sqrt a - \sqrt b }} - \frac{{3\sqrt {ab} }}{{a\sqrt a - b\sqrt b }}} \right):\left( {\frac{{a - b}}{{a + \sqrt {ab} + b}}} \right)\\
= \left( {\frac{{a - \sqrt {ab} + b + 3\sqrt {ab} }}{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}} \right).\left( {\frac{{a + \sqrt {ab} + b - 3\sqrt {ab} }}{{\left( {\sqrt a - \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}} \right):\left( {\frac{{a - b}}{{a + \sqrt {ab} + b}}} \right)\\
= \frac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}.\frac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{\left( {\sqrt a - \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}:\frac{{a - b}}{{a + \sqrt {ab} + b}}\\
= \frac{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}{{\left( {a - \sqrt {ab} + b} \right)\left( {a + \sqrt {ab} + b} \right)}}:\frac{{a - b}}{{a + \sqrt {ab} + b}}\\
= \frac{{a - b}}{{\left( {a - \sqrt {ab} + b} \right)\left( {a + \sqrt {ab} + b} \right)}}.\frac{{a + \sqrt {ab} + b}}{{a - b}}\\
= \frac{1}{{a - \sqrt {ab} + b}}
\end{array}\)
