Question

CMR: \(\dfrac{n^4+3n^3+2n^2+6n-2}{n^2+2}\). có giá trị là 1 số nguyên với n€N

Answer 1

Ta có: \(\dfrac{n^4+3n^3+2n^2+6n-2}{n^2+2}\)

\(=\dfrac{\left(n^4+2n^2\right)+\left(3n^3+6n\right)-2}{n^2+2}\)

\(=\dfrac{n^2\left(n^2+2\right)+3n\left(n^2+2\right)-2}{n^2+2}\)

\(=\dfrac{n^2\left(n^2+2\right)+3n\left(n^2+2\right)}{n^2+2}-\dfrac{2}{n^2+2}\)

Ta thấy: \(n^2\left(n^2+2\right)⋮n^2+2;3n\left(n^2+2\right)⋮n^2+2\)

\(\Rightarrow n^2\left(n^2+2\right)+3n\left(n^2+2\right)⋮n^2+2\)

\(\dfrac{2}{n^2+2}=\dfrac{4}{2n^2+4}=\dfrac{4}{2\left(n^2+2\right)}\)

do \(4⋮2\Rightarrow4⋮2\left(n^2+2\right)\) (đoạn này mk ko chắc chắn cho lắm ~.~)

Khi đó: \(n^2\left(n^2+2\right)+3n\left(n^2+2\right)-2⋮n^2+2\)

-> ĐPCM.