Đáp án: $B = \dfrac{{ - 10 - 6\sqrt 5 }}{5}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x > 0;x\# 4\\
B = \dfrac{{2 + \sqrt x }}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
= \dfrac{{\left( {2 + \sqrt x } \right)\left( {\sqrt x - 2} \right) - \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 - x}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 4}}{{x - 2\sqrt x }}\\
x = \dfrac{{3 - \sqrt 5 }}{2}\left( {tmdk} \right)\\
= \dfrac{{6 - 2\sqrt 5 }}{4}\\
= \dfrac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{{2^2}}}\\
\Leftrightarrow \sqrt x = \dfrac{{\sqrt 5 - 1}}{2}\\
\Leftrightarrow B = \dfrac{4}{{\dfrac{{3 - \sqrt 5 }}{2} - 2.\dfrac{{\sqrt 5 - 1}}{2}}}\\
= \dfrac{8}{{3 - \sqrt 5 - 2\sqrt 5 + 2}}\\
= \dfrac{8}{{5 - 3\sqrt 5 }}\\
= \dfrac{{8\left( {5 + 3\sqrt 5 } \right)}}{{ - 20}}\\
= \dfrac{{ - 10 - 6\sqrt 5 }}{5}
\end{array}$
