Question

[Đặng Thị Thanh Tâm]

tìm số nguyên x,y sao cho

x²-xy=6x-5y-8

x²+y²=4x-6y+12

Answer 1

[tran pham bao thy]

x²-xy=6x-5y-8

x.x-x.y-6.x+5.y= -8

Answer 2

[ʚDʉү_²ƙ⁶ɞ‏]

b, \(x^2+y^2=4x-6y+12\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=-1\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^2=-1\)

Sai đề nha bạn!!