Đáp án:
$\begin{array}{l}
a)\left( {\dfrac{1}{{\sqrt 3 - 2}} - \dfrac{1}{{\sqrt 3 + 2}}} \right).\dfrac{{2 - \sqrt 2 }}{{1 - \sqrt 2 }}\\
= \dfrac{{\sqrt 3 + 2 - \sqrt 3 + 2}}{{\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}}.\dfrac{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }}\\
= \dfrac{4}{{3 - 4}}.\left( { - \sqrt 2 } \right)\\
= 4\sqrt 2 \\
b)Dkxd:x \ne \dfrac{1}{2}\\
A = 3x + \dfrac{{\sqrt {4{x^2} - 4x + 1} }}{{1 - 2x}}\\
= 3x + \dfrac{{\sqrt {{{\left( {1 - 2x} \right)}^2}} }}{{1 - 2x}}\\
= 3x + \dfrac{{\left| {1 - 2x} \right|}}{{1 - 2x}}\\
Khi:x = - 2\left( {tmdk} \right)\\
\Leftrightarrow A = 3.\left( { - 2} \right) + \dfrac{{\left| {1 - 2.\left( { - 2} \right)} \right|}}{{1 - 2.\left( { - 2} \right)}}\\
= - 6 + \dfrac{5}{5}\\
= - 6 + 1\\
= - 5
\end{array}$
