\(2n^2+2n-3=2n\left(n+1\right)-3⋮n+1\)
\(\Rightarrow3⋮n+1\)
Mà \(n\inℤ\Rightarrow n+1\inℤ\)
\(\Rightarrow n+1\in\left\{-1;1;3;-3\right\}\)
\(\Rightarrow n\in\left\{-2;0;-4;2\right\}\)
Gọi f( x ) = 2n2 + 2n - 3
g( x ) = n + 1
Cho g( x ) = 0
\(\Leftrightarrow\)n + 1 = 0
\(\Leftrightarrow\)n = - 1
\(\Leftrightarrow\)f( x ) = 2 . ( - 1 )2 - 2 . ( - 1 ) - 3 = - 3
Để f( x ) \(⋮\)g( x ) thì n +1 \(\inƯ_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)
Ta lập bảng :
| \(n+1\) | \(-1\) | \(1\) | \(-3\) | \(3\) |
| \(n\) | \(-2\) | \(0\) | \(-4\) | \(2\) |
\(Vậy:n\in\left\{-4;-2;0;2\right\}\)
Ta có:
\(2n^2+2n-3=2n\left(n+1\right)-3\)
Để n nguyên thì \(-3⋮\left(n+1\right)\)
\(\Rightarrow n+1\inƯ\left(-3\right)\)
\(\Rightarrow n+1\in\left\{-1,1,3,-3\right\}\)( Vì n nguyên nên n+1 nguyên)
\(\Rightarrow n\in\left\{-2,0,2,-4\right\}\)
Vậy ... Tích cho mk nha :)) :-*
