Đáp án:a/ 4a
b/ $F = - 4\sqrt 6 + 12$
Giải thích các bước giải:
$F = \left( {\frac{{\sqrt a + 1}}{{\sqrt a - 1}} - \frac{{\sqrt a - 1}}{{\sqrt a + 1}} + 4\sqrt a } \right).\left( {\sqrt a - \frac{1}{{\sqrt a }}} \right)$
Đk: $\{ _{a \ne 1}^{a > 0}$
a/ $\begin{array}{l}
F = \left( {\frac{{\sqrt a + 1}}{{\sqrt a - 1}} - \frac{{\sqrt a - 1}}{{\sqrt a + 1}} + 4\sqrt a } \right).\left( {\sqrt a - \frac{1}{{\sqrt a }}} \right)\\
= \left( {\frac{{{{(\sqrt a + 1)}^2} - {{(\sqrt a - 1)}^2} + 4\sqrt a (\sqrt a + 1)(\sqrt a - 1)}}{{(\sqrt a - 1)(\sqrt a + 1)}}} \right).\left( {\frac{{a - 1}}{{\sqrt a }}} \right)\\
= \left( {\frac{{(\sqrt a + 1 + \sqrt a - 1)(\sqrt a + 1 - \sqrt a + 1) + 4\sqrt a (a - 1)}}{{a - 1}}} \right).\frac{{a - 1}}{{\sqrt a }}\\
= \frac{{4\sqrt a + 4\sqrt a (a - 1)}}{{a - 1}}.\frac{{a - 1}}{{\sqrt a }}\\
= \frac{{4\sqrt a (1 + a - 1)}}{{\sqrt a }} = 4a
\end{array}$
b/ Thay $a = \frac{{\sqrt 6 }}{{2 + \sqrt 6 }}$ ta được:
$F = 4a = \frac{{4\sqrt 6 }}{{2 + \sqrt 6 }} = \frac{{4\sqrt 6 .(2 - \sqrt 6 )}}{{4 - 6}} = \frac{{8\sqrt 6 - 24}}{{ - 2}} = - 4\sqrt 6 + 12$
