Question

[Nguyễn Ngân]

Tìm x,y nguyên biết:

x.y+x -2y+1 =0

Answer 1

[Sakura Akari]

xy + x - 2y + 1 = 0

xy + x - 2y - 2 = -3

(xy - 2y) + (x - 2) = -3

y(x - 2) + (x - 2) = -3

(x - 2)(y + 1) = -3

Vì x, y nguyên \(\Rightarrow\) x - 2 \(\in\)Ư(-3) = {1; -1; 3; -3}

Lập bảng giá trị

x - 2	1	-1	3	-3
x	3	1	5	-1
y + 1	-3	3	-1	1
y	-4	2	-2	0

Vậy (x; y) \(\in\) { (3; -4) , (1; 2) , (5; -2) , (-1; 0) }

Answer 2

[Nguyễn Thanh Hằng]

\(xy+x-2y+1=0\)

\(\Leftrightarrow x\left(y+1\right)-2y+1=0\)

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

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

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

Ta có các trường hợp :

+) \(\left\{{}\begin{matrix}y+1=1\\x-2=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=5\end{matrix}\right.\)

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

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

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

Vậy....