Ta có: \(\left(2n+3\right)⋮\left(2n+1\right)\)
\(\Rightarrow\left(2n+1+2\right)⋮\left(2n+1\right)\)
Vì \(\left(2n+1\right)⋮\left(2n+1\right)\) nên để \(\left(2n+1+2\right)⋮\left(2n+1\right)\) thì \(2⋮\left(2n+1\right)\)
\(\Rightarrow2n+1\inƯ\left(2\right)\)
\(\Rightarrow2n+1\in\left\{1;2;-1;-2\right\}\)
\(\Rightarrow2n\in\left\{0;1;-2;-3\right\}\)
\(\Rightarrow n\in\left\{0;\frac{1}{2};-1;-\frac{3}{2}\right\}\)
\(2n+3⋮2n+1\)
\(2n+3=2n+1+2⋮2n+1\)
mà \(2n+1⋮2n+1\)
\(\Rightarrow2⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(2\right)\left\{1;2\right\}\)
| 2n + 1 | 1 | 2 |
| 2n + 1 | 0 | \(n\notinℕ\) |
Vậy \(n=0\)
sai thì cho mk xl nha!!!
\(\left(2n+3\right)⋮2n+1\)
\(\Rightarrow\left(2n+1+2\right)⋮2n+1\)
\(\text{Vì}\left(2n+1\right)⋮\left(2n+1\right)\text{nên}2⋮\left(2n+1\right)\)
\(\Rightarrow\left(2n+1\right)\inƯ\left(2\right)\)
\(\RightarrowƯ\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow\left(2n+1\right)\in\left\{1;2\right\}\)
\(\Rightarrow\orbr{\begin{cases}2n+1=1\\2n+1=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}n=0\\n=\frac{1}{2}\end{cases}}\)
