Ta có \(2n+5⋮2n+1\)
\(\Rightarrow\left(2n+1\right)+4⋮2n+1\)
\(\Rightarrow4⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow2n\in\left\{-5;-3;-2;0;1;3\right\}\)
\(\Rightarrow n\in\left\{\frac{-5}{2};\frac{-3}{2};-1;0;\frac{1}{2};\frac{3}{2}\right\}\)
Mà n nguyên
\(\Rightarrow n\in\left\{-1;0\right\}\)
Vậy \(n\in\left\{-1;0\right\}\)
@@ Học tốt !!
#Chiyuki Fujito
\(\left(2n+5\right)⋮\left(2n+1\right)\)
\(\left(2n+1\right)+4⋮\left(2n+1\right)\)mà\(\left(2n+1\right)⋮\left(2n+1\right)\)
\(\Rightarrow4⋮\left(2n+1\right)\Rightarrow2n+1\inƯ\left(4\right)=\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow2n+1\in\left\{-1;1;-2;2;-4;4\right\}\)
Vì \(n\inℤ\)nên ta có bảng sau:
| 2n + 1 | -1 | 1 | -2 | 2 | -4 | 4 |
| n | -1 | 0 | \(\frac{-3}{2}\) | \(\frac{1}{2}\) | \(\frac{-5}{2}\) | \(\frac{3}{2}\) |
mà \(n\inℤ\Rightarrow n\in\left\{-1;0\right\}\)
Vậy \(n\in\left\{-1;0\right\}\)
Chúc bạn học tốt!!!
