Đáp án:
$\begin{array}{l}
19)B = {\left( {\dfrac{{\sqrt a }}{2} - \dfrac{1}{{2\sqrt a }}} \right)^2}.\left( {\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} - \dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}} \right)\\
= {\left( {\dfrac{{\sqrt a .\sqrt a - 1}}{{2\sqrt a }}} \right)^2}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2} - {{\left( {\sqrt a + 1} \right)}^2}}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}\\
= \dfrac{{{{\left( {a - 1} \right)}^2}}}{{4a}}.\dfrac{{ - 4\sqrt a }}{{a - 1}}\\
= \dfrac{{ - \left( {a - 1} \right)}}{{\sqrt a }}\\
= \dfrac{{1 - a}}{{\sqrt a }}\\
B4)\\
1)\left( {5\sqrt {\dfrac{1}{5}} + \dfrac{1}{2}\sqrt {20} - \dfrac{5}{2}\sqrt {\dfrac{4}{5}} + \sqrt 5 } \right):2\sqrt 5 \\
= \left( {\sqrt 5 + \dfrac{1}{2}.2\sqrt 5 - \dfrac{5}{2}.\dfrac{2}{{\sqrt 5 }} + \sqrt 5 } \right):2\sqrt 5 \\
= \left( {\sqrt 5 + \sqrt 5 - \sqrt 5 + \sqrt 5 } \right):2\sqrt 5 \\
= 2\sqrt 5 :2\sqrt 5 \\
= 1\\
2)\left( {\dfrac{{2\sqrt 3 - \sqrt 6 }}{{\sqrt 8 - 2}} - \dfrac{{\sqrt {216} }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\
= \left( {\dfrac{{\sqrt 6 .\left( {\sqrt 2 - 1} \right)}}{{2\left( {\sqrt 2 - 1} \right)}} - \dfrac{{6\sqrt 6 }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\
= \left( {\dfrac{{\sqrt 6 }}{2} - 2\sqrt 6 } \right).\dfrac{1}{{\sqrt 6 }}\\
= \dfrac{1}{2} - 2\\
= - \dfrac{3}{2} = - 1,5\\
3)\left( {\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 - \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }} + \dfrac{{\sqrt 5 \left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( { - \sqrt 7 - \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {\sqrt 7 + \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {7 - 5} \right)\\
= - 2
\end{array}$
