Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = \dfrac{{a + 4\sqrt a + 4}}{{\sqrt a + 2}} + \dfrac{{4 - a}}{{\sqrt a - 2}}\\
= \dfrac{{{{\left( {\sqrt a + 2} \right)}^2}}}{{\sqrt a + 2}} + \dfrac{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}}{{\sqrt a - 2}}\\
= \sqrt a + 2 - \left( {2 + \sqrt a } \right)\\
= 0\\
2,\\
A = \dfrac{{a + b + 2\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a + \sqrt b }}\\
= \left( {\sqrt a + \sqrt b } \right) - \left( {\sqrt a - \sqrt b } \right)\\
= 2\sqrt b \\
3,\\
A = \left( {\dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }} - \sqrt {ab} } \right).{\left( {\dfrac{{\sqrt a + \sqrt b }}{{a - b}}} \right)^2}\\
= \left( {\dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}{{\sqrt a + \sqrt b }} - \sqrt {ab} } \right).{\left( {\dfrac{{\sqrt a + \sqrt b }}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}} \right)^2}\\
= \left( {\left( {a - \sqrt {ab} + b} \right) - \sqrt {ab} } \right).{\left( {\dfrac{1}{{\sqrt a - \sqrt b }}} \right)^2}\\
= \left( {a - 2\sqrt {ab} + b} \right).\dfrac{1}{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}\\
= {\left( {\sqrt a - \sqrt b } \right)^2}.\dfrac{1}{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}\\
= 1
\end{array}\)
