a) \(\frac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\frac{1}{{\sqrt a + \sqrt b }} = a - b\) với a > 0; b > 0
Xét vế trái, ta có:
\(\begin{array}{l}VT = \frac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\frac{1}{{\sqrt a + \sqrt b }}\\ = \frac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\left( {\sqrt a + \sqrt b } \right)\\ = \left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)\\ = a - b = VP\end{array}\)
Vậy \(\frac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\frac{1}{{\sqrt a + \sqrt b }} = a - b\)
b) \(\left( {1 + \frac{{a + \sqrt a }}{{\sqrt a + 1}}} \right)\left( {1 - \frac{{a - \sqrt a }}{{\sqrt a - 1}}} \right) = 1 - a\) với a \( \ge \) 0 và a \( \ne \)1
Xét vế trái ta có:
\(\left( {1 + \frac{{a + \sqrt a }}{{\sqrt a + 1}}} \right)\left( {1 - \frac{{a - \sqrt a }}{{\sqrt a - 1}}} \right) = \left( {1 + \frac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}}} \right)\left( {1 - \frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\sqrt a - 1}}} \right)\)
\( = \left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right) = 1 - {\left( {\sqrt a } \right)^2} = 1 - a\) = VP.
Vậy \(\left( {1 + \frac{{a + \sqrt a }}{{\sqrt a + 1}}} \right)\left( {1 - \frac{{a - \sqrt a }}{{\sqrt a - 1}}} \right) = 1 - a\)