A=x^2 -y^2 +6xy +9
A=(x+3y)^2 -10y^2 +9
A=(x+3y)^2 -(10y^2-9)
TH1 10y^2 -9<0
A >0 moi x,y => ko co nhan tu
TH2
10y^2-9>=0
A=[x+3y-can(10y^2-9)][x+3y+can(10y^2-9)]
mình sửa đề xíu
\(x^2-y^2+6x+9\)
\(=x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
=\(\left(x+3-y\right)\left(x+3+y\right)\)