|
|
Nhận xét: $a.b > 0 \Rightarrow a, b > 0$ hoặc $a, b < 0$ và $a + b \neq 0$
$1 + \frac{1}{4}{\left( {\sqrt {\frac{a}{b}} - \sqrt {\frac{b}{a}} } \right)^2} = 1 +
\frac{1}{4}\left( {\frac{a}{b} + \frac{b}{a} - 2} \right) = \frac{{{{\left( {a + b}
\right)}^2}}}{{4ab}}$
$ \Rightarrow A = \frac{{2\sqrt {ab} }}{{a + b}}.\frac{{\sqrt {{{\left( {a + b}
\right)}^2}} }}{{2\sqrt {ab} }} = \frac{{\left| {a + b} \right|}}{{a + b}} = \left[ \begin{array}{l}
1\,\,\,\,khi\,\,\,\,\,\,a,b > 0\\
- 1\,\,\,\,\,khi\,\,\,\,{\rm{a}},b < 0
\end{array} \right.$
|