$ĐK:3x\geq y$
Xét $(1)\Leftrightarrow 9x^2-6xy+y^2=7x^4-6x^2y^2+2x^3y+y^2$
$\Leftrightarrow 9x^2-6xy-7x^4+6x^2y^2-2x^3y=0$
$\Leftrightarrow 6x^2y^2-2y(x^3+3x)-7x^4+9x^2=0(2)$
+) $x=0\Rightarrow ........$
+)$x\neq 0\Rightarrow .............$
Coi $(2)$ là phương trình bậc $2$ ẩn $y,$ có:
$\Delta 'y=x^6+6x^4+9x^2+42x^4-54x^2=x^6-6x^4+9x^2=(x^3-3x)^2$
$\Rightarrow ............$
Chị lm típ nhak! ^^