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Giải
Ta có : $x = \frac{1}{2}\left( {\sqrt {\frac{{ - a}}{{ - b}} + } \sqrt {\frac{{ - b}}{{ - a}}} }
\right) = - \frac{{a + b}}{{2\sqrt {ab} }},\sqrt {{x^2} - 1} = \frac{{\left| {a - b}
\right|}}{{2\sqrt {ab} }}$
$\Rightarrow $$E = \frac{{ - 2\left| {a - b} \right|}}{{a + b - \left| {a - b} \right|}} = \left[
\begin{array}{l}
\frac{{a\left( {a - b} \right)}}{{2\sqrt {ab} }}\,\,\,\,khi \,\,\,\,\,\,\,a \ge b\\a - b\,\,\,\,\,\,\,khi\,\,\,\,\,\,\,\,a < b\end{array} \right.$
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Đăng bài 10-05-12 11:22 AM
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