a.
\(x^2-3y^2+2xy-2x-10y+4=0\)
\(\Leftrightarrow\left(x^2-xy-3x\right)+\left(3xy-3y^2-9y\right)+\left(x-y-3\right)+7=0\)
\(\Leftrightarrow x\left(x-y-3\right)+3y\left(x-y-3\right)+\left(x-y-3\right)=-7\)
\(\Leftrightarrow\left(x-y-3\right)\left(x+3y+1\right)=-7\)
Ta có bảng:
| x-y-3 | -7 | -1 | 1 | 7 |
| x+3y+1 | 1 | 7 | -7 | -1 |
| x | -3 | 3 | 1 | 7 |
| y | 1 | 1 | -3 | -3 |
Vậy \(\left(x;y\right)=\left(-3;1\right);\left(3;1\right);\left(1;-3\right);\left(7;-3\right)\)
b.
\(\Leftrightarrow2x^2-3xy-2y^2+6x-2y+4=5\)
\(\Leftrightarrow\left(2x^2-4xy+4x\right)+\left(xy-2y^2+2y\right)+\left(2x-4y+4\right)=5\)
\(\Leftrightarrow2x\left(x-2y+2\right)+y\left(x-2y+2\right)+2\left(x-2y+2\right)=5\)
\(\Leftrightarrow\left(x-2y+2\right)\left(2x+y+2\right)=5\)
Ta có bảng sau:
| x-2y+2 | -5 | -1 | 1 | 5 |
| 2x+y+2 | -1 | -5 | 5 | 1 |
| x | -13/5 | -17/5 | 1 | 1/5 |
| y | 11/5 | -1/5 | 1 | -7/5 |
Vậy pt có đúng 1 cặp nghiệm \(\left(x;y\right)=\left(1;1\right)\)
