1) \(\left(x-1\right)\left(y-1\right)=5\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=1.5=5.1=\left(-1\right).\left(-5\right)=\left(-5\right).\left(-1\right)\)
Ta có bảng sau:
\(x-1\) | \(1\) | \(5\) | \(-1\) | \(-5\) |
\(y-1\) | \(5\) | \(1\) | \(-5\) | \(-1\) |
\(x\) | \(2\) | \(6\) | \(0\) | \(-4\) |
\(y\) | \(6\) | \(2\) | \(-4\) | \(0\) |
Vậy \(\left(x;y\right)=\left(2;6\right);\left(6;2\right);\left(0;-4\right);\left(-4;0\right)\)
2) \(x\left(y+2\right)=-8\)
\(\Leftrightarrow x\left(y+2\right)=2.\left(-4\right)=\left(-4\right).2=\left(-2\right).4=4.\left(-2\right)=1.\left(-8\right)=\left(-8\right).1=\left(-1\right).8=8.\left(-1\right)\)
Ta có bảng sau:
\(x\) | \(2\) | \(-4\) | \(-2\) | \(4\) | \(1\) | \(-8\) | \(-1\) | \(8\) |
\(y+2\) | \(-4\) | \(2\) | \(4\) | \(-2\) | \(-8\) | \(1\) | \(8\) | \(-1\) |
\(y\) | \(-6\) | \(0\) | \(2\) | \(-4\) | \(-10\) | \(-1\) | \(6\) | \(-3\) |
Vậy ...