Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
mx + 2y = m + 1\\
2x + my = - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2mx + 4y = 2m + 2\\
2mx + {m^2}y = - 2m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4y - {m^2}y = 2m + 2 + 2m\\
2x = - 2 - my
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {4 - {m^2}} \right).y = 4m + 2\\
2x = - 2 - my
\end{array} \right.\\
\Leftrightarrow 4 - {m^2}\# 0\\
\Leftrightarrow {m^2}\# 4\\
\Leftrightarrow m\# 2;m\# - 2\\
Vậy\,m\# 2;m\# - 2\\
b)Dkxd:x \ge 0;x\# 4\\
A = \left( {\dfrac{1}{{\sqrt x - 2}} - \dfrac{1}{{2 + \sqrt x }}} \right):\dfrac{4}{{{x^2} - 4x + 4}}\\
= \dfrac{{2 + \sqrt x - \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{{{\left( {x - 2} \right)}^2}}}{4}\\
= \dfrac{4}{{x - 2}}.\dfrac{{{{\left( {x - 2} \right)}^2}}}{4}\\
= x - 2
\end{array}$
