T
tuan.hocmai


1> [tex]\left\{ \begin{array}{l} x(3x+2y)(x+1)= 12 \\ x^2 + 2y +4x - 8 =0\end{array} \right.[/tex]
2> [tex]\left\{ \begin{array}{l} (2x+y)^2 - 5(4x^2-y^2) + 6(2x-y)^2 =0 \\ 2x+y + \frac{1}{2x-y}=3\end{array} \right.[/tex]
2> [tex]\left\{ \begin{array}{l} (2x+y)^2 - 5(4x^2-y^2) + 6(2x-y)^2 =0 \\ 2x+y + \frac{1}{2x-y}=3\end{array} \right.[/tex]