Question

tìm x,  thuộc Z biết:

xy+x+y=1

Answer 1

xy + x + y = 1

=> x(y + 1) + (y + 1) = 2

=> (x + 1)(y + 1) = 2

=> x + 1; y + 1 \(\in\)Ư(2) = {1; -1; 2; -2}

Lập bảng :

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

Vậy ...

Answer 2

\(xy+x+y=1\)

\(\Rightarrow x\left(y+1\right)+\left(y+1\right)=2\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=2\)

\(\Rightarrow x+1;y+1\inƯ\left(2\right)=\left\{\mp1;\mp2\right\}\)

Ta có bảng xét :

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