Question

Tìm số nguyên x ; y biết xy = x-y

Answer 1

\(xy=x-y\)

\(\Leftrightarrow xy-\left(x-y\right)=0\)

\(\Leftrightarrow xy-x+y=0\)

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

\(\Leftrightarrow\left(y-1\right)\left(x+1\right)=-1=-1.1=1.\left(-1\right)\)

Lập bảng:

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

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

Answer 2

\(xy=x-y\)

\(\Rightarrow xy-x+y=0\)

\(\Rightarrow xy-x+y-1=-1\)

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

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

Vì \(x;y\in Z\)nên xét bảng:

x + 1	1	-1
y - 1	-1	1
x	0	-2
y	0	2

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