Ta có:\(A=\dfrac{2n^2+4n+5}{n+1}\)
\(=\dfrac{\left(2n^2+2n\right)+\left(2n+2\right)+3}{n+1}\)
\(=\dfrac{2n.\left(n+1\right)+2.\left(n+1\right)+3}{n+1}\)
Để \(A\) nhận giá trị nguyên thì:
\(n+1\in\left\{\pm1;\pm3\right\}\)
\(TH1:\)\(n+1=1\Rightarrow n=0\)
\(TH2:n+1=-1\Rightarrow n=-2\)
\(TH3:n+1=3\Rightarrow n=2\)
\(TH4:n+1=-3\Rightarrow n=-4\)
Vậy \(n\in\left\{0;-2;2;-4\right\}\)
A=\(\dfrac{\left(2n^2+2n\right)+\left(2n+2\right)+3}{n+1}\)
A=\(\dfrac{2n.\left(n+1\right)+2.\left(n+1\right)+3}{n+1}\)
Để C nhận giá trị nguyên thì n+1 thuộc (1;-1;3;-3)
TH1 : n + 1 = 1 => n= 0
TH2 : n + 1= -1 => n =-2
TH3 : n + 1 = 3 => n = 2
TH4 : n + 1 = -3 => n=-4
n thuộc (0;2;-2;-4)
