Đáp án:
$\begin{array}{l}
1)A = \left( {\dfrac{1}{{\sqrt a - 2}} - \dfrac{2}{{2 - 2\sqrt a }}} \right).\left( {\dfrac{{a - 3\sqrt a + 2}}{{\sqrt a - 1}} + 1} \right)\\
= \left( {\dfrac{1}{{\sqrt a - 2}} + \dfrac{1}{{\sqrt a - 1}}} \right).\left( {\dfrac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a - 2} \right)}}{{\sqrt a - 1}} + 1} \right)\\
= \dfrac{{\sqrt a - 1 + \sqrt a - 2}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}.\left( {\sqrt a - 2 + 1} \right)\\
= \dfrac{{2\sqrt a - 3}}{{\sqrt a - 2}}\\
2)\left\{ \begin{array}{l}
3x - 2y = 5\\
x + 3y = - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3x - 2y = 5\\
3x + 9y = - 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
11y = - 11\\
x = - 2 - 3y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = - 1\\
x = 1
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {1; - 1} \right)\\
3)a)m = 2\\
\Leftrightarrow {x^2} - 2.\left( {2 + 1} \right).x + {2^2} + 4 = 0\\
\Leftrightarrow {x^2} - 6x + 8 = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0\\
\Leftrightarrow x = 2;x = 4\\
Vậy\,x = 2;x = 4
\end{array}$
