\(x-xy+y=2x-y\)
\(\Rightarrow-x-xy+2y=0\)
\(\Rightarrow-x\left(1+y\right)=-2y\)
\(\Rightarrow x=\dfrac{2y}{y+1}=\dfrac{2\left(y+1\right)-2}{y+1}=2-\dfrac{2}{y+1}\left(y\ne-1;x\ne2\right)\)
-Ta có: x,y là số nguyên.
\(\Rightarrow2⋮\left(y+1\right)\)
\(\Rightarrow y+1\inƯ\left(2\right)\)
| y+1 | 1 | -1 | 2 | -2 |
| y | 0(n) | -2(n) | 1(n) | -3(n) |
\(y=0\Rightarrow x=2-\dfrac{2}{0+1}=0\left(n\right)\)
\(y=-2\Rightarrow x=2-\dfrac{2}{-2+1}=4\left(n\right)\)
\(y=1\Rightarrow x=2-\dfrac{2}{1+1}=1\left(n\right)\)
\(y=-3\Rightarrow x=2-\dfrac{2}{-3+1}=3\left(n\right)\)
-Vậy các cặp số (x,y) là (0,0); (4,-2); (1,1) ;(3;-3)
Đúng 4
Bình luận (0)
