a. 2xy + 3x - 2y - 8 = 0
\(\left(2xy+3x\right)-2y-3-5=0\)
\(x\left(2y+3\right)-\left(2y+3\right)=5\)
\(\left(2y+3\right)\left(x-1\right)=5\)
Với \(x,y\in Z\) nên 2y + 3 và \(x-1\in Z\)
Do đó 2y + 3 và \(x-1\inƯ\left(5\right)\)
Ta có 4 TH:
TH1: \(\left\{{}\begin{matrix}x-1=1\\2y+3=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\) ( t/m )
TH2: \(\left\{{}\begin{matrix}x-1=5\\2y+3=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=-1\end{matrix}\right.\) ( t/m )
TH3: \(\left\{{}\begin{matrix}x-1=-1\\2y+3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=-4\end{matrix}\right.\) ( t/m )
TH4: \(\left\{{}\begin{matrix}x-1=5\\2y+3=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\) ( t/m)
Vậy các cặp ( x;y ) là : ( 2;1 ), ( 6;-1 ), ( 0;-4 ), ( -4;-2 )
\(2xy+3x-2y-8=0\)
\(\Rightarrow2\left(2xy+3x\right)-2\left(2y+8\right)=0\)
\(\Rightarrow4xy+6x-4y-8=0\)
\(\Rightarrow4xy+6x-4y-6-2=0\)
\(\Rightarrow4y\left(x-1\right)+6\left(x-1\right)-2=0\)
\(\Rightarrow\left(4y+6\right)\left(x-1\right)-2=0\)
\(\Rightarrow\left(4y+6\right)\left(x-1\right)=2\)
Xét ước 2
\(\left(4y+6\right).\left(x-1\right)=2\)a. 2xy + 3x - 2y - 8 = 0
=> \(2.\left(2xy+3x\right)-2.\left(2y+8\right)=0\)
=> \(4xy+6x-4y-8=0\)
=> \(4y.\left(x-1\right)+6.\left(x-1\right)-2=0\)
=> \(\left(4y+6\right).\left(x-1\right)-2=0\)
=> \(\left(4y+6\right).\left(x-1\right)=2\)
Xét ước của 2
Vậy cặp số \(\left(x;y\right)\) là : \(\left(2;1\right),\left(6;-1\right),\left(0;-4\right),\left(-4;-2\right)\)
