xy + 4x + y = 3
x . (y + 4) + y = 3
x . (y + 4) + (y + 4) = 7
(y + 4) . (x + 1) = 7
TC x,y e Z và y + 4 ; x + 1 e U(7) = {+1; +7}
TC bảng sau
y + 4 | -1 | -7 | 1 | 7 |
x + 1 | -7 | -1 | 7 | 1 |
y | -5 | -11 | -3 | 3 |
x | -8 | -2 | 6 | 0 |
Vậy (x,y) e (-5; -8) ; (-11; -2) ; (-3; 6) ; (3; 0)
sai mong bạn thông cảm
HT
xy+4x+y=3xy+4x+y=3
⇒x(y+4)+(y+4)=3+4⇒x(y+4)+(y+4)=3+4
⇒(x+1)(y+4)=7⇒(x+1)(y+4)=7
⇒(x+1);(y+4)∈Ư(7)={±1;±7}⇒(x+1);(y+4)∈Ư(7)={±1;±7}
Ta có các trường hợp sau
TH1:\hept{x+1=1y+4=7⇔\hept{x=0y=3TH1:\hept{x+1=1y+4=7⇔\hept{x=0y=3 TH2:\hept{x+1=−1y+4=−7⇔\hept{x=−2y=−11TH2:\hept{x+1=−1y+4=−7⇔\hept{x=−2y=−11
TH3:\hept{x+1=7y+4=1⇔\hept{x=6y=−3TH3:\hept{x+1=7y+4=1⇔\hept{x=6y=−3 TH4:\hept{x+1=−7y+4=−1⇔\hept{x=−8y=−5TH4:\hept{x+1=−7y+4=−1⇔\hept{x=−8y=−5
Vậy(x;y)∈{(0;3);(−2;−11);(6;−3);(−8;−5)}