2n-3 chia hết cho n+1
=>2n+2-5 chia hết cho n+1
=>2(n+1)-5 chia hết cho n+1
Mà 2(n+1) chia hết cho n+1=>5 chia hết cho n+1
=>n+1 thuộc Ư(5)={1;-1;5;-5}
TH1:n+1=1,=>n=0 thuộc Z
TH2:n+1=-1=>n=-2 thuộc Z
TH3:n+1=5=>n=4 thuộc Z
TH4:n+1=-5=>n=-6 thuộc Z
Vậy n thuộc {0;-2;4;-6}
Chúc Bạn Học Tốt !!!
\(2n-3⋮n+1\)
\(\Rightarrow2n+2-5⋮n+1\)
\(\Rightarrow2\left(n+1\right)-5⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=-1\Rightarrow n=-2\\n+1=5\Rightarrow n=4\\n+1=-5\Rightarrow n=-6\end{matrix}\right.\)
Ta có :
\(2n-3⋮n+1\)
Mà \(n+1⋮n+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2n-3⋮n+1\\2n+2⋮n+1\end{matrix}\right.\)
\(\Leftrightarrow5⋮n+1\)
Vì \(n\in Z\Leftrightarrow n+1\in Z;n+1\inƯ\left(5\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}n+1=1\\n+1=5\\n+1=-1\\n+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=0\\n=4\\n=-2\\n=-6\end{matrix}\right.\)
Vậy .....
