a, xy+3x-7y=21. => x(y+3) - 7(y+3) = 0. => (x-7)(y+3)=0. => x=7 , và với mọi y. Hoặc y=3 với mọi x
\(a)\)
\(xy+3x-7y=21\)
\(xy+3x-7y-21=0\)
\(x\left(y+3\right)-7\left(y+3\right)=0\)
\(\Rightarrow x=7;y=\left(-3\right)\)
Vậy ...
\(b)\)
\(xy+3x-2y=11\)
\(xy+3x-2y-6=5\)
\(x\left(y+3\right)-2\left(y+3\right)=5\)
\(\left(x-2\right)\left(y+3\right)=5\)
\(\Rightarrow x-2;y+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có:
| \(x-2\) | \(\left(-1\right)\) | \(\left(-5\right)\) | \(1\) | \(5\) |
| \(x\) | \(1\) | \(\left(-3\right)\) | \(3\) | \(7\) |
| \(y+3\) | \(\left(-5\right)\) | \(\left(-1\right)\) | \(5\) | \(1\) |
| \(y\) | \(\left(-8\right)\) | \(\left(-4\right)\) | \(2\) | \(2\) |
Vậy ...