Question

n^2+6n-3 chia hết cho n+3

Answer 1

\(n^2+6n-3⋮n+3\)

\(\Rightarrow n\left(n+3\right)+n-3⋮n+3\)

\(\Rightarrow n+3-6⋮n+3\)

\(\Rightarrow-6⋮n+3\Rightarrow n+3\inƯ\left(-6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

Bạn tự làm tiếp nha

Answer 2

\(Bổ\ sung\ đề\ bài : Tìm\ n\in Z\)

\(ĐK:n\ne-3\)

\(n^2+6n-3=\left(n^2+3n\right)+\left(3n+9\right)-12\\ =n\left(n+3\right)+3\left(n+3\right)-12\\ =\left(n+3\right)^2-12\)

\(Theo\ đề\ bài :\) \(n^2+6n-3⋮\left(n+3\right)\)

\(=>\left(n+3\right)^2-12⋮\left(n+3\right)\\ =>12⋮\left(n+3\right)\)

\(=>n+3\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

\(=>n\in\left\{-4;-5;-6;-7;-9;-15;-2;-1;0;1;3;9\right\}\left(TMDK\right)\)