\(\left(x-1\right)\left(2x-y\right)=3\)
\(\Rightarrow x-1\inƯ\left(3\right)\)
\(x-1\in\left\{-1,-3,1,3\right\}\)
\(x-1\in N\Rightarrow x-1\in\left\{1,3\right\}\)
Tương tự, \(2x-y\in\left\{1,3\right\}\)
\(\left(x-1\right)\left(2x-y\right)=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1;2x-y=3\\x-1=3;2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2.2-y=3\Rightarrow-y=-1\Rightarrow y=1\\2.4-y=3\Rightarrow-y=-5\Rightarrow y=5\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\) và \(\left\{{}\begin{matrix}y=1\\y=5\end{matrix}\right.\) ư
Câu đúng nà:
\(\left(x-1\right)\left(2x-5\right)=3\)
\(\Rightarrow x-1\inƯ\left(3\right)\Rightarrow x-1\in\left\{-1,-3,1,3\right\}\)
Mà \(x-1\in N\Rightarrow x-1\left\{1,3\right\}\)
Tương tự, \(2x-y\in\left\{1,3\right\}\)
\(\left(x-1\right)\left(2x-y\right)=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1;2x-y=3\\x-1=3;2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2.2-y=3\Rightarrow y=1\\2.4-y=1\Rightarrow y=7\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}x=2;y=1\\x=4;y=7\end{matrix}\right.\)
