xy + 3x-2y=11
<=> x(y+3)-2(y+3)=5
<=>(x-2)(y+3)=5
suy ra (x-2) và (y+3) là các ước nguyên của 5.
Th1. x-2=1 <=>x=3
.......y+3=5 <=> y=2
Th2 x-2=-1 <=> x=1
.......y+3=-5 <=> y= -8
Th3. x-2=5 <=> x=7
.......y+3=1 <=> y= -2
Th4. x-2= -5 <=> x= -3
.......y+3= -1 <=> y= -4
Vậy (x,y) = (3, 2); (1, -8); (7, -2); (-3, -4)
\(xy+2x+y+11=0\)
\(\Rightarrow x.\left(y+2\right)+y+2+9=0\)
\(\Rightarrow\left(y+2\right).\left(x+1\right)=-9\)
\(\Rightarrow y+2\) và \(x+1\inƯ\left(-9\right)\)
Ta xét các trường hợp sau:
\(TH1:\left\{\begin{matrix}y+2=1\\x+1=-9\end{matrix}\right.\Rightarrow\left\{\begin{matrix}y=-1\\x=-10\end{matrix}\right.\)
\(TH2:\left\{\begin{matrix}y+2=3\\x+1=-3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}y=1\\x=-4\end{matrix}\right.\)
\(TH3\left\{\begin{matrix}y+2=9\\x+1=-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}y=7\\x=-2\end{matrix}\right.\)
\(TH4:\left\{\begin{matrix}y+2=-3\\x+1=3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}y=-5\\x=2\end{matrix}\right.\)
Vậy \(\left(y;x\right)=\left(-1;-10\right);\left(1;4\right);\left(7;-2\right)\left(-5;2\right)\)
