Ta có: $\left\{\begin{array}{1} x^{2}+y^{2}=10 \\ 6x^{2}-12x+8=(x+y)(10-xy) \end{array}\right.$
$\implies \left\{\begin{array}{1} x^{2}+y^{2}=10 \\ 6x^{2}-12x+8=(x+y)(x^2 + y^2 - xy) \end{array}\right.$
$\implies \left\{\begin{array}{1} x^{2}+y^{2}=10 \\ 6x^{2}-12x+8=x^3 + y^3 \end{array}\right.$
$\implies \left\{\begin{array}{1} x^{2}+y^{2}=10 \\ -x^3 + 6x^{2}-12x+8 = y^3 \end{array}\right.$
$\implies \left\{\begin{array}{1} x^{2}+y^{2}=10 \\ (2-x)^3 = y^3 \end{array}\right.$
$\implies \left\{\begin{array}{1} x^{2}+y^{2}=10 \\ 2-x = y \end{array}\right.$
$\implies \left\{\begin{array}{1} x^{2}+(2-x)^{2}=10 \\ 2-x = y \end{array}\right.$
$\implies \left\{\begin{array}{1} 2x^{2} - 4x - 6 = 0 \\ 2-x = y \end{array}\right.$
$\implies \left\{\begin{array}{1}
\left[\begin{array}{1}
x = -1
\\
x = 3
\end{array}\right.
\\
2-x = y
\end{array}\right.$
+) $x = -1 \implies y = 3$
+) $x = 3 \implies y = -1$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
