a)
$\begin{array}{l} \left\{ \begin{array}{l} x - 2y + 2 = 0\\ 2y - {x^2} = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2y = {x^2}\left( 1 \right)\\ x - 2y + 2 = 0\left( 2 \right) \end{array} \right.\\ \left( 1 \right) \to \left( 2 \right) \Rightarrow x - {x^2} + 2 = 0\\ \Leftrightarrow {x^2} - x - 2 = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 1 \Rightarrow y = \frac{{{x^2}}}{2} = \frac{1}{2}\\ x = 2 \Rightarrow y = \frac{{{x^2}}}{2} = \frac{4}{2} = 2 \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( { - 1;\frac{1}{2}} \right),\left( {2;2} \right) \end{array}$
b)
$\begin{array}{l} \left\{ \begin{array}{l} x + y + xy = 7\\ {x^2} + {y^2} + xy = 13 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x + y + xy = 7\\ {\left( {x + y} \right)^2} - xy = 13 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} S + P = 7\left( 1 \right)\\ {S^2} - P = 13\left( 2 \right) \end{array} \right.\left( {S = x + y,P = xy} \right)\left( {{S^2} - 4P \ge 0} \right)\\ \left( 1 \right) \to \left( 2 \right) \Rightarrow {S^2} - \left( {7 - S} \right) = 13\\ \Leftrightarrow {S^2} + S - 20 = 0\\ \Leftrightarrow \left( {S + 5} \right)\left( {S - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} S = - 5 \Rightarrow P = 12\left( L \right)\\ S = 4 \Rightarrow P = 3\left( {TM} \right) \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} S = 4\\ P = 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + y = 4\\ xy = 3 \end{array} \right.\\ \Rightarrow x,y\,là\,nghiệm\,của\,PT\,{t^2} - 4t + 3 = 0\\ \Leftrightarrow \left( {t - 1} \right)\left( {t - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t = 3 \end{array} \right. \Rightarrow \left( {x;y} \right) = \left( {1;3} \right),\left( {3;1} \right) \end{array}$
