Question

tìm cặp số nguyên x và y

2x-xy+y=-1

Answer 1

\(2x-xy+y=-1\)

\(\Leftrightarrow2x-xy+y-2=-3\)

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

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

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

Answer 2

2x - xy + y = -1
<=> 2x - xy = -1 - y
<=> x(2 - y) = 2 - y - 3
<=> x(2 - y) - (2 - y) = -3
<=> (x - 1)(2 - y) = -3
Do x; y \(\in Z\Rightarrow x-1;2-y\in Z\)
Mà (x - 1)(2 - y) = -3
=> x - 1; 2 - y \(\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Nếu \(\left\{{}\begin{matrix}x-1=1\\2-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-1\\2-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=3\\2-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-3\\2-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\) (thỏa mãn)

Vậy các cặp (x; y) thỏa mãn là (2; 5); (0; -1); (4; 3); (-2; 1)
​ @nguyễn phương anh