Đáp án:
$\begin{array}{l}
a)\left| {x - 2} \right| = 3x + 2\\
Dk:3x + 2 \ge 0 \Rightarrow x \ge - \dfrac{2}{3}\\
\Rightarrow \left[ \begin{array}{l}
x - 2 = 3x + 2\\
x - 2 = - 3x - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = - 4\\
4x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 2\left( {ktm} \right)\\
x = 0\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 0\\
b)Dkxd:x \ge 1\\
\sqrt {{x^2} + 3} = x - 1\\
\Rightarrow {x^2} + 3 = {x^2} - 2x + 1\\
\Rightarrow 2x = - 2\\
\Rightarrow x = - 1\left( {ktm} \right)
\end{array}$
Vậy pt vô nghiệm.
$\begin{array}{l}
c)\left\{ \begin{array}{l}
2x + 3y = 7\\
- 2x + y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x + 3y - 2x + y = 7 + 1\\
- 2x + y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4y = 8\\
2x = y - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = 2\\
2x = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2};y = 2
\end{array}$
