Đáp án:
$\begin{array}{l}
a)\\
E = \dfrac{{\sqrt a + \sqrt b - 1}}{{a + \sqrt {ab} }} + \dfrac{{\sqrt a - \sqrt b }}{{2\sqrt {ab} }}.\left( {\dfrac{{\sqrt b }}{{a - \sqrt {ab} }} + \dfrac{{\sqrt b }}{{a + \sqrt {ab} }}} \right)\\
= \dfrac{{\sqrt a + \sqrt b - 1}}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}\\
+ \dfrac{{\sqrt a - \sqrt b }}{{2\sqrt {ab} }}.\sqrt b .\dfrac{{\sqrt a + \sqrt b + \sqrt a - \sqrt b }}{{\sqrt a \left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{{\sqrt a + \sqrt b - 1}}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}} + \dfrac{1}{{2\sqrt a }}.\dfrac{{2\sqrt a }}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{{\sqrt a + \sqrt b - 1}}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}} + \dfrac{1}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{{\sqrt a + \sqrt b }}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{1}{{\sqrt a }}\\
b)\\
F = 2ab\sqrt {\dfrac{{ - 1}}{{ab}}} + 3b\sqrt {\dfrac{{ - a}}{b}} + a\sqrt {\dfrac{{ - b}}{a}} \\
= - 2\sqrt {{a^2}{b^2}.\dfrac{{ - 1}}{{ab}}} + 3\sqrt b .\sqrt { - a} + \sqrt a .\sqrt { - b} \\
= - 2\sqrt { - ab} + 3\sqrt { - ab} + \sqrt { - ab} \\
= 2\sqrt { - ab}
\end{array}$
