Question

Tìm STN x,y để xy+x-y=4

Answer 1

\(xy+x-y=x\left(y+1\right)-y=4\)
\(x\left(y+1\right)-y-1=3\)
\(x\left(y+1\right)-\left(y+1\right)=3\)
\(\left(x-1\right)\left(y+1\right)=3\)
\(=>\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.=>\left[{}\begin{matrix}y+1=3\\y+1=-3\\y+1=1\\y+1=-1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2\\y=-4\\y=0\\y=-2\end{matrix}\right.\)

Answer 2

\(\text{x(y+1)-y=4}\)

\(\text{x(y+1)-y-1=3}\)

\(\text{x(y+1)-(y+1)=3}\)

\(\text{(x-1)(y+1)=3}\)

=> \(\left(x-1\right);\left(y+1\right)\in\left(Ư\right)3=\left\{1;3;-1;-3\right\}\)

==> bảng:

x-1	1	3	-1	-3
y+1	3	1	-3	-1
x	2	4	0	-2
y	2	0	-4	-2

=>  (x;y) ∈ { (2;2);(4;0);(0;-4);(-2;-2)}