Ta có:\(m-1⋮2m+1\)
\(\Leftrightarrow4\left(m-1\right)⋮2m+1\)
\(\Leftrightarrow4m-4⋮2m+1\)
\(\Leftrightarrow4m+2-6⋮2m+1\)
\(\Leftrightarrow2\left(2m+1\right)-6⋮2m+1\)
\(\Leftrightarrow6⋮2m+1\)
\(\Leftrightarrow2m+1\inƯ\left(6\right)\)
Mà 2m+1 là số lẻ \(\forall m\in Z\)
\(\Rightarrow2m+1\in\left\{-3;-1;1;3\right\}\)
\(\Rightarrow2m\in\left\{-4;-2;0;2\right\}\)
\(\Rightarrow m\in\left\{-2;-1;0;1\right\}\)
Vậy có 4 số nguyên m
theo bài ra ta có:
\(m-1⋮2m+1\\ \Rightarrow2m-2⋮2m+1\\ \Rightarrow2m+1-3⋮2m+1\\ \Rightarrow3⋮2m+1\\ \Rightarrow2m+1\inƯ_{\left(3\right)}=\left\{1;-1;3;-3\right\}\)
ta có bảng sau:
| 2m+1 | 1 | -1 | 3 | -3 |
| 2m | 0 | -2 | 2 | -4 |
| m | 0 | -1 | 1 | -2 |
vậy m ={0; -1; 1; -2}
