\(xy-y+2x-5=0\\ \Leftrightarrow xy-y+2x-2-3=0\\ \Leftrightarrow\\ \Leftrightarrow\left(xy-y\right)+\left(2x-2\right)=3\\ \Leftrightarrow y\left(x-1\right)+2\left(x-1\right)=3\\ \Leftrightarrow\left(y+2\right)\left(x-1\right)=3\\ \Leftrightarrow\left[{}\begin{matrix}\left(y+2\right)\left(x-1\right)=3\cdot1\\\left(y+2\right)\left(x-1\right)=1\cdot3\\\left(y+2\right)\left(x-1\right)=\left(-1\right)\left(-3\right)\\\left(y+2\right)\left(x-1\right)=\left(-3\right)\left(-1\right)\end{matrix}\right.\)
Ta lập bảng giá trị:
| \(x-1\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(y+2\) | \(-1\) | \(-3\) | \(3\) | \(1\) |
| \(x\) | \(-2\) | \(0\) | \(2\) | \(4\) |
| \(y\) | \(-3\) | \(-5\) | \(1\) | \(-1\) |
Vậy cắp số \(\left\{x;y\right\}\) tương ứng là \(\left\{-2;-3\right\};\left\{0;-5\right\};\left\{2;1\right\};\left\{4;-1\right\}\)
lộn câu c)|x-1|=0
=>x-1=0
=>x=0+1
=>x=1
tick tớ nha