Question

Tìm cặp số nguyên (x,y) thoả mãn x + y + xy =2.

Answer 1

Ta có:

\(x+y+xy=2\)

\(\Leftrightarrow x+y+xy+1=3\)

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

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

Mà \(3=1.3=3.1=\left(-1\right).\left(-3\right)=\left(-3\right).\left(-1\right)\)

Ta có bảng sau:

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

Vậy \(\left(x;y\right)=\left(2;0\right);\left(0;2\right);\left(-2;-4\right);\left(-4;-2\right)\)