\(2x-5y+5xy=13\)
\(x\left(2+5y\right)-\left(2+5y\right)=11\)
\(\left(x+1\right)\left(2+5y\right)=11\)
Ta có : \(11=1.11=\left(-1\right)\left(-11\right)\)
Ta có bảng sau :
| \(x+1\) | \(-11\) | \(-1\) | \(1\) | \(11\) |
| \(2+5y\) | \(-1\) | \(-11\) | \(11\) | \(1\) |
| \(x\) | \(-12\) | \(-2\) | \(0\) | \(10\) |
| \(y\) | \(\dfrac{-3}{5}\) | \(\dfrac{-13}{5}\) | \(\dfrac{9}{5}\) | \(\dfrac{-1}{5}\) |
Vậy \(\left(x;y\right)\in\left\{\left(-12;\dfrac{-3}{5}\right);\left(-2;\dfrac{-13}{5}\right);\left(0;\dfrac{9}{5}\right);\left(10;\dfrac{-1}{5}\right)\right\}\)
Bạn quên là nó \(\ge0\le\)
ak