Rút gọn các biểu thức (biết a> 0, b > 0):
a) \(\sqrt {\frac{a}{b}} + \sqrt {\frac{b}{a}} - \frac{{\sqrt {ab} }}{a}\);
b) \(\left( {a - 2\sqrt {\frac{b}{a}} } \right)\left( {a + \frac{2}{a}\sqrt {ab} } \right)\).
Dựa vào: \(\frac{{\sqrt a }}{{\sqrt b }} = \frac{{\sqrt a .\sqrt b }}{{{{\left( {\sqrt b } \right)}^2}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
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\(\sqrt {\frac{a}{b}} = \sqrt {\frac{{ab}}{{{b^2}}}} = \frac{{\sqrt {ab} }}{b}(a \ge 0,b > 0)\)
a) \(\sqrt {\frac{a}{b}} + \sqrt {\frac{b}{a}} - \frac{{\sqrt {ab} }}{a}\)
\(= \sqrt {\frac{{ab}}{{{b^2}}}} + \sqrt {\frac{{ab}}{{{a^2}}}} - \frac{{\sqrt {ab} }}{a} \\= \frac{{\sqrt {ab} }}{b} + \frac{{\sqrt {ab} }}{a} - \frac{{\sqrt {ab} }}{a} \\= \frac{{\sqrt {ab} }}{b}.\)
b) \(\left( {a - 2\sqrt {\frac{b}{a}} } \right)\left( {a + \frac{2}{a}\sqrt {ab} } \right) \)
\(= \left( {a - 2\sqrt {\frac{b}{a}} } \right)\left( {a + 2\sqrt {\frac{b}{a}} } \right) \\= {a^2} - \frac{{4b}}{a}.\)