Ta có:
\(xy-x+2y=3\)
\(\Rightarrow xy-x+2y-2=1\)
\(\Rightarrow x.\left(y-1\right)+2.\left(y-1\right)=1\)
\(\Rightarrow\left(y-1\right).\left(x+2\right)=1\)
\(\Rightarrow y-1;x+2\inƯ\left(1\right)\)
\(\Rightarrow y-1;x+2\in\left\{-1;1\right\}\)
Ta có bảng sau:
| y-1 | -1 | 1 |
| x+2 | -1 | 1 |
| y | 0 | 2 |
| x | -3 | -1 |
Vậy \(\left(x;y\right)\in\left\{\left(-3;0\right);\left(-1;2\right)\right\}\)
Chúc bạn học tốt!!!
Theo đề bài ta có:
\(xy-x+2y=3\)
\(\Leftrightarrow xy-x+2y-2=3-2\)
\(\Leftrightarrow xy-x+2y-2=1\)
\(\Leftrightarrow x.y-x.1+2.y-2.1=1\)
\(\Leftrightarrow x\left(y-1\right)+2\left(y-1\right)=1\)
\(\Leftrightarrow\left(x+2\right)\left(y-1\right)=1\)
\(\Leftrightarrow x+2;y-1\inƯ\left(1\right)\)
\(Ư\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x+2=1\Rightarrow x=-1\)
\(y-1=-1\Rightarrow y=0\)
\(\Leftrightarrow x+2=-1\Rightarrow x=-3\)
\(y-1=1\Rightarrow y=2\)
Theo đề bài ta có:
xy−x+2y=3xy−x+2y=3
⇔xy−x+2y−2=3−2⇔xy−x+2y−2=3−2
⇔xy−x+2y−2=1⇔xy−x+2y−2=1
⇔x.y−x.1+2.y−2.1=1⇔x.y−x.1+2.y−2.1=1
⇔x(y−1)+2(y−1)=1⇔x(y−1)+2(y−1)=1
⇔(x+2)(y−1)=1⇔(x+2)(y−1)=1
⇔x+2;y−1∈Ư(1)⇔x+2;y−1∈Ư(1)
Ư(1)={±1}
Ta có bảng sau:
|
y-1 |
-1 |
1 |
|
x+2 |
-1 |
1 |
|
y |
0 |
2 |
|
x |
-3 |
-1 |
Vậy (x;y)∈{(−3;0);(−1;2)}(x;y)∈{(−3;0);(−1;2)}
