Question

[Chii]

xy-3x+2y-11=0 tìm x,y thuộc Z

Answer 1

[Kiệt Nguyễn]

                            Giải

Theo đề bài, ta có: \(xy-3x+2y-11=0\)

\(\Leftrightarrow x\left(y-3\right)+2y-6=5\)

\(\Leftrightarrow x\left(y-3\right)+2\left(y-3\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(y-3\right)=5\)

\(\Leftrightarrow\hept{\begin{cases}x+2\\y-3\end{cases}}\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Lập bảng:

\(x+2\)	\(1\)	\(-1\)	\(5\)	\(-5\)
\(y-3\)	\(5\)	\(-5\)	\(1\)	\(-1\)
\(x\)	\(-1\)	\(-3\)	\(3\)	\(-7\)
\(y\)	\(8\)	\(-2\)	\(4\)	\(2\)

Vậy \(\left(x,y\right)\in\left\{\left(-1,8\right);\left(-3,-2\right);\left(3,4\right);\left(-7,2\right)\right\}\)