\(xy+12=x-y\)
\(\Rightarrow xy-x+y=-12\)
\(\Rightarrow x\left(y-1\right)+\left(y-1\right)=-13\)
\(\Rightarrow\left(y-1\right)\left(x+1\right)=-13\)
\(\Rightarrow\left(y-1\right)\left(x+1\right)=-1.13=-13.1=1.\left(-13\right)=13.\left(-1\right)\)
Đến đây bn lập bảng lak tìm ra x,y
\(xy+12=x-y\)
\(\Rightarrow xy-x+y=-12\)
\(\Rightarrow x\left(y-1\right)+\left(y-1\right)=-12-1\)
\(\Rightarrow\left(x+1\right)\left(y-1\right)=-13\)
\(\Rightarrow\left(x+1\right)\left(1-y\right)=13\)
\(\Rightarrow\left(x+1\right);\left(1-y\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
Xét bảng
x+1 | 1 | -1 | 13 | -13 |
1-y | 13 | -13 | 1 | -1 |
x | 0 | -2 | 12 | -14 |
y | -12 | 14 | 0 | 2 |
Vậy......................................
Ta có
xy+12=x-y
=>x(y-1)+(y-1)=-13
Đến đây ok rồi chứ