Question

[Wendy Marvell]

tìm các số nguyên x, y sao cho : x^2-xy=6x-5y-8

Answer 1

[Pham Van Hung]

\(x^2-xy=6x-5y-8\)

\(\Rightarrow x^2-xy-6x+5y+8=0\)

\(\Rightarrow\left(x^2-xy-x\right)-\left(5x-5y-5\right)+3=0\)

\(\Rightarrow x\left(x-y-1\right)-5\left(x-y-1\right)=-3\)

\(\Rightarrow\left(x-y-5\right)\left(x-1\right)=-3\)

Từ đó bạn tìm ước thì ra kết quả.Chúc bạn học tốt.

Answer 2

[Đinh quang hiệp]

đặt \(x-y=k\)

\(x^2-xy=6x-5y-8\Rightarrow x\left(x-y\right)=x+\left(5x-5y\right)-8\Rightarrow xk=x+5\left(x-y\right)-8\)

\(\Rightarrow xk=x+5k-8\Rightarrow xk=x+5k-5-3\Rightarrow xk-x-5k+5=-3\)

\(\Rightarrow x\left(k-1\right)-5\left(k-1\right)=3\Rightarrow\left(x-5\right)\left(k-1\right)=3\Rightarrow x-5;k-1\inƯ\left(-3\right)=+-1;+-3\)

nếu \(x-5=1\Rightarrow x=6\)thì \(k-1=-3\Rightarrow k=-2\Rightarrow y=x-k=6-\left(-2\right)=8\)

nếu \(x-5=3\Rightarrow x=8\)thì \(k-1=-1\Rightarrow k=0\Rightarrow y=x-k=8-0=8\)

nếu \(x-5=-1\Rightarrow x=4\)thì \(k-1=3\Rightarrow k=4\Rightarrow y=x-k=4-4==0\)

nếu \(x-5=-3\Rightarrow x=2\)thì \(k-1=1\Rightarrow k=2\Rightarrow y=x-k=2-2=0\)

vậy (x;y)=(6;8) (8;8) (4;0) (2;0)