1. $\begin{cases}2x+y=\frac{3}{x^2} \\ 2y+x=\frac{3}{y^2} \end{cases}$
Tru 2 ve cua pt cho nhau ta dk
$(x-y)(1+\frac{3(x+y)}{x^2.y^2})=0$
$TH1: x=y $ de dang giai dk $x=y=1$
$TH2: 1+\frac{3x+3y}{x^2y^2}=0<=>-3(x+y)=x^2y^2$
Cong hai ve cua hpt ban dau co
$3(x+y)=\frac{3x^2+3y^2}{x^2y^2}<=>3(x+y)=\frac{3x^2+3y^2}{-3(x+y)}$
$<=>-3(x+y)^2=x^2+y^2<=>2x^2+3xy+2y^2=0 (vo nghiem) $