Ta có : \(\frac{x+1}{2}=\frac{3}{y-2}\Rightarrow\left(x+1\right)\left(y-2\right)=2.3=6\)
- Vì : \(x\in Z\Rightarrow x+1\in Z\)
\(y\in Z\Rightarrow y-2\in Z\)
Phân tích 6 thành tích hai số nguyên ta có :
6 = 1.6 = (-1) (-6)
\(\Rightarrow\left(x+1\right)\left(y-2\right)=1.6=\left(-1\right).\left(-6\right)\)
+) Nếu \(\left\{\begin{matrix}x+1=1\\y-2=6\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=1-1=0\\y=6+2=8\end{matrix}\right.\)
+) Nếu \(\left\{\begin{matrix}x+1=6\\y-2=1\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=6-1=5\\y=1+2=3\end{matrix}\right.\)
+) Nếu \(\left\{\begin{matrix}x+1=-6\\y-2=-1\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=-6-1=-7\\y=-1+2=1\end{matrix}\right.\)
+) Nếu \(\left\{\begin{matrix}x+1=-1\\y-2=-6\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=-1-1=-2\\y=-6+2=-4\end{matrix}\right.\)
Vậy ...
Do \(\frac{x+1}{2}\)=\(\frac{3}{y-2}\)
\(\Rightarrow\)(x+1).(y-2)=2.3
\(\Rightarrow\) (x+1).(y-2)=6
\(\Rightarrow\)x+1 và y-2 \(\in\) Ư(6)
\(\Rightarrow\)x+1 và y-2 \(\in\){ \(\pm\)1;\(\pm\)2;\(\pm\)3;\(\pm\)6}
*) Nếu x+1=1 thì y-2=6
\(\Rightarrow\)x=0 thì y=8
*) Nếu x+1=-1 thì y-2=-6
\(\Rightarrow\)x = -2 thì y=-4
*) Nếu x+1=2 thì y-2=3
\(\Rightarrow\)x=1 thì y=5
*) Nếu x+1=-2 thì y-2=-3
\(\Rightarrow\)x =-3 thì y =-1
*) Nếu x+1=6 thì y-2=1
\(\Rightarrow\)x=5 thì y=3
*) Nếu x+1=-6 thì y-2=-1
\(\Rightarrow\)x=-7 thì y=1
*) Nếu x+1=3 thì y-2=2
\(\Rightarrow\)x=2 thì y=4
*) Nếu x+1=-3 thì y-2=-2
\(\Rightarrow\)x=-4 thì y=0
