Đáp án:
Giải thích các bước giải:
$\Leftrightarrow \left\{ \begin{array}{l} {x^3} - {y^3} - 3({x^2} + {y^2}) - 9(x - y) + 22 = 0\\ {x^2} + {y^2} = \frac{1}{2} + x - y \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} ({x^3} - {y^3}) - 12(x - y) + \frac{{41}}{2} = 0\\ {x^2} + {y^2} = \frac{1}{2} + x - y \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {(x - y)^3} + 3xy(x - y) - 12(x - y) + \frac{{41}}{2} = 0\\ {(x - y)^2} + 2xy = \frac{1}{2} + x - y \end{array} \right.$
Tới đấy đặt a=x-y và b=xy có:
$\Leftrightarrow \left\{ \begin{array}{l} {a^3} + 3ab - 12a + \frac{{41}}{2} = 0\\ {a^2} - a + 2b = \frac{1}{2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \frac{{ - 1}}{2}{a^3} + \frac{3}{2}{a^2} - \frac{{45}}{4}a + \frac{{41}}{2} = 0\\ b = \frac{1}{4} + \frac{a}{2} - \frac{{{a^2}}}{2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 2\\ b = - \frac{3}{4} \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x = 2 + y\\ 2y + {y^2} + \frac{3}{4} = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2 + y\\ \left[ \begin{array}{l} y = - \frac{1}{2}\\ y = - \frac{3}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{3}{2}\\ y = - \frac{1}{2} \end{array} \right.\\ \left\{ \begin{array}{l} x = \frac{1}{2}\\ y = - \frac{3}{2} \end{array} \right. \end{array} \right. \end{array} \right.$
