xy-3x-3y=2
=>x(y-3)-3y=2
<=>x(y-3)-3y-9=11
<=>x(y-3)-3(y-3)=11
<=>(x-3)(y-3)=11
x-3 | 11 | 1 | -11 | -1 |
y-3 | 1 | 11 | -1 | -11 |
x | 14 | 4 | -8 | 2 |
y | 4 | 14 | 2 | -8 |
Vậy (x;y)=(14;4), (4:14), (-8;2) ,(2; -8)
nhớ k cho mik nhé cảm ơn
ta co xy-3x-3y=2
=>x*(y-3)-3y+9=2+9
=>x(y-3)-3(y-3)=11
=>(x-3)(y-3)=11
=>x-3 va y-3 thuoc uoc cua 11
Ma uoc cua 11={+-1;+-11}
Ta co bang sau:
x-3 1 11 -1 -11
y-3 11 1 -11 -1
x 4 14 2 -8
y 14 4 -8 2
Vay (x;y) thuoc {(4;14);(14;4);(2;-8);(-8;2)}