<=>(x-1)y+2x=3
=>(x-1)y+2y-3=0
=>x=1
=>x=-2
<=> (x-1)y+2x=3
=> (x-1)y+2y-3=0
=> x = 1
=> x = -2
ta co :x.(y+2)-y=3
x.y+x.2-y-2=3-2
x.y-y+x.2-2=1
y.(x-1)+2.(x-1)=1
(x-1).(y+2)=1
ma 1=1.1=-1.-1
khi x-1=1thi x=2
y+2=1 thi y=-1
khi x-1=-1 thi x=0
y+2=-1 thi x=-3
\(x.\left(y+2\right)-y=3\)
\(\Rightarrow x.\left(y+2\right)-\left(y+2\right)=5\)
\(\Rightarrow\left(x-1\right)\left(y+2\right)=5\)
Bn kẻ bảng rồi tính nốt nha@_@