Vì \(\left(2x+11\right)\) \(⋮\) \(\left(5x+1\right)\) mà \(\left(5x+1\right)\) \(⋮\) \(\left(5x+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}5\left(2x+11\right)⋮\left(5x+1\right)\\2\left(5x+1\right)⋮\left(5x+1\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(10x+55\right)⋮\left(5x+1\right)\\\left(10x+2\right)⋮\left(5x+1\right)\end{matrix}\right.\)
\(\Rightarrow\left(10x+55\right)-\left(10x+2\right)⋮\left(5x+1\right)\)
\(\Rightarrow53⋮\left(5x+1\right)\)
\(\Rightarrow\left(5x+1\right)\inƯ\left(53\right)\)
\(\Rightarrow\left(5x+1\right)\in\left\{\pm1;\pm53\right\}\)
Ta có bảng sau:
| \(5x+1\) | \(-53\) | \(-1\) | \(1\) | \(53\) |
| \(5x\) | \(-54\) | \(-2\) | \(0\) | \(52\) |
| \(x\) |
\(\dfrac{-54}{5}\) (loại) |
\(\dfrac{-2}{5}\) (loại) |
\(0\) (TM) |
\(\dfrac{-52}{5}\) (loại) |
Vậy \(x=0\) thì \(\left(2x+11\right)⋮\left(5x+1\right)\).
