Lời giải:
Vì \(x,y\in\mathbb{Z}\Rightarrow x-2; xy-1\in\mathbb{Z} \)
Do đó, với \((x-2)(xy-1)=5\) có thể xảy ra các TH sau:
TH1: \(\left\{\begin{matrix} x-2=1\\ xy-1=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3\\ xy=6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3\\ y=2\end{matrix}\right.\) (chọn)
TH2: \(\left\{\begin{matrix} x-2=5\\ xy-1=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=7\\ xy=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=7\\ y=\frac{2}{7}\end{matrix}\right.\) (loại)
TH3: \(\left\{\begin{matrix} x-2=-1\\ xy-1=-5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ xy=-4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=-4\end{matrix}\right.\) (chọn)
TH4: \(\left\{\begin{matrix} x-2=-5\\ xy-1=-1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-3\\ xy=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-3\\ y=0\end{matrix}\right.\) (chọn)
Vậy \((x,y)\in \left\{(3;2); (1;-4); (-3;0)\right\}\)
Tìm các số nguyên x và y biết rằng ( x - 2 ) (xy -1) = 5
Vì ( x - 2 ) (xy -1) \(\in\) Z
Ta cho :
(x-2)= 1 hoặc (x-2)= 5
\(\left\{{}\begin{matrix}x-2=1\\x-2=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Ta cho
(xy-1)=5 hoặc (xy-1)= 1
thay x=3
Ta được:
(3.y-1)=5
=> 3y= 6
=> y= 2
...................Thay vào và tính tương tự...................
Theo đề bài ta có:
x;y \(\in\) Z
\(\Rightarrow\) x - 2 ; xy - 1 \(\in\) Z
Ta có các trường hợp
TH1:
\(\left\{{}\begin{matrix}x-2=1\\xy-1=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1+2\\xy=5+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\xy=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) ( thỏa mãn )
TH2:
\(\left\{{}\begin{matrix}x-2=5\\xy-1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5+2\\xy=1+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=7\\xy=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=\dfrac{2}{7}\end{matrix}\right.\) ( loại )
TH3:
\(\left\{{}\begin{matrix}x-2=-1\\xy-1=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\left(-1\right)+2\\xy=\left(-5\right)+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\xy=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-4\end{matrix}\right.\) ( thỏa mãn )
TH4:
\(\left\{{}\begin{matrix}x-2=-5\\xy-1=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\left(-5\right)+2\\xy=\left(-1\right)+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\xy=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\) ( thỏa mãn )
Vậy (x;y) \(\in\) { ( 3;2 ) ; ( 1;-4 ) ; ( -3;0 ) }
