Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \left( {\dfrac{{2\sqrt a + \sqrt b }}{{\sqrt {ab} + 2\sqrt a - \sqrt b - 2}}} \right) - \left( {\dfrac{{2\sqrt a - \sqrt b }}{{\sqrt {ab} + 2\sqrt a + \sqrt b + 2}}} \right)\\
= \dfrac{{2\sqrt a + \sqrt b }}{{\sqrt a \left( {\sqrt b + 2} \right) - \left( {\sqrt b + 2} \right)}} - \dfrac{{2\sqrt a - \sqrt b }}{{\sqrt a \left( {\sqrt b + 2} \right) + \left( {\sqrt b + 2} \right)}}\\
= \dfrac{{2\sqrt a + \sqrt b }}{{\left( {\sqrt a - 1} \right)\left( {\sqrt b + 2} \right)}} - \dfrac{{2\sqrt a - \sqrt b }}{{\left( {\sqrt a + 1} \right)\left( {\sqrt b + 2} \right)}}\\
= \dfrac{{\left( {2\sqrt a + \sqrt b } \right)\left( {\sqrt a + 1} \right) - \left( {2\sqrt a - \sqrt b } \right)\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2a + 2\sqrt a + \sqrt {ab} + \sqrt b - \left( {2a - 2\sqrt a - \sqrt {ab} + \sqrt b } \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{4\sqrt a + 2\sqrt {ab} }}{{\left( {\sqrt a - 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2\sqrt a \left( {\sqrt b + 2} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2\sqrt a }}{{a - 1}}
\end{array}\)
