Đặt \(A=3x-xy+y=12\)
\(\Leftrightarrow3x-xy+y-12=0\)
\(\Leftrightarrow x\left(3-y\right)-\left(3-y\right)-9=0\)
\(\Leftrightarrow\left(x-1\right)\left(3-y\right)=9\)
Ta có:\(9=3x3=\left(-3\right)x\left(-3\right)\)
Do đó ta có bảng sau:
| x-1 | 3 | -3 |
| 3-y | 3 | -3 |
| x | 4 | -2 |
| y | 0 | 6 |
Vậy cặp ( x; y ) TM là:(4;0)(-2;6)
3x - xy + y = 12
3x - y . ( x - 1 ) = 12
3x - 3 - y . ( x - 1 ) = 9
3 . ( x - 1 ) - y . ( x - 1 ) = 9
( 3 - y ) . ( x - 1 ) = 9
Ta có bảng sau :
| 3-y | 1 | 9 | -1 | -9 | 3 | -3 |
| x-1 | 9 | 1 | -9 | -1 | 3 | -3 |
| y | 2 | -6 | 4 | 12 | 0 | 6 |
| x | 10 | 2 | -8 | 0 | 4 | -2 |
Vậy ( x ; y ) = ( 2 ; 10 ) = ( -6 ; 2 ) = ( 4 ; -8 ) = ( 12 ; 0 ) = ( 0 ; 4 ) = ( 6 ; -2 )
3x-xy-y-12=0
3x-xy=12+y
(3-y)x=12+y
(-3+y).x=12+y
=>(-3+15).2,25=12+15=27
vay x=2,25;y=15.
lam lieu do ko dung dau
