Question

Tìm  x ,y nguyên  sao cho : xy + 3x -y = 6

Answer 1

Ta có : 

\(xy+3x-y=6\)

\(\Rightarrow x\left(y+3\right)-y-3=6-3\)

\(\Rightarrow x\left(y+3\right)-\left(y+3\right)=3\)

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

Do \(x;y\in Z\)

\(\Rightarrow x-1;y+3\in Z\)

\(\Rightarrow x-1;y+3\inƯ\left(3\right)\)

Nên ta có bảng sau :

\(x-1\)	\(1\)	\(3\)	\(-1\)	\(-3\)
\(y+3\)	\(3\)	\(1\)	\(-3\)	\(-1\)
\(x\)	\(2\)	\(4\)	\(0\)	\(-2\)
\(y\)	\(0\)	\(-2\)	\(-6\)	\(-4\)

Vậy  \(\left(x;y\right)\in\left\{\left(2,0\right);\left(4,-2\right);\left(0;-6\right);\left(-2;-4\right)\right\}\)

~ Ủng hộ nhé 

P/s : Đúng nha