Question

tìm các cặp số nguyên x,y thỏa mãn biết xy+4x+y=3

Answer 1

Ta có: xy+4x+y=3

\(\Leftrightarrow\) x(y+4)+(y+4)=3+4=7

\(\Leftrightarrow\) (y+4).(x+1)=7

Ta có: (y+4).(x+1)=1.7=-1.-7

\(\Rightarrow\left[{}\begin{matrix}y+4=1;x+1=7\\y+4=7;x+1=1\\y+4=-1;x+1=-7\\y+4=-7;x+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}y=1-4=-3;x=7-1=6\\y=7-4=3;x=1-1=0\\y=-1-4=-5;x=-7-1=-8\\y=-7-4=-11;x=-1-1=-2\end{matrix}\right.\)

Vậy..........................................