Câu hỏi của Nguyen Thuy Linh

Question

chứng minh rằng với mọi số nguyên n thì:

S=$\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1$ chia hết cho 5

Lightning Farron · Accepted Answer

$S=\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1$

$=2n\left(n^2-3n-1\right)+\left(n^2-3n-1\right)-2n^3+1$

$=2n^3-6n^2-2n+n^2-3n-1-2n^3+1$

$=\left(2n^3-2n^3\right)-\left(6n^2-n^2\right)-\left(2n+3n\right)-1+1$

$=-5n^2-5n=-5n\left(n+1\right)⋮5$Câu trả lời: $S=\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1$

$=2n\left(n^2-3n-1\right)+\left(n^2-3n-1\right)-2n^3+1$

$=2n^3-6n^2-2n+n^2-3n-1-2n^3+1$

$=\left(2n^3-2n^3\right)-\left(6n^2-n^2\right)-\left(2n+3n\right)-1+1$

$=-5n^2-5n=-5n\left(n+1\right)⋮5$

Lưu Ngọc Hải Đông · Answer

$S=\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1$

$=2n^3-6n^2-2n+n^2-3n-1-2n^3+1$

$=-5n^2-5n=-5n\left(n+1\right)⋮5$

Vậy $\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1⋮5$Câu trả lời: $S=\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1$

$=2n^3-6n^2-2n+n^2-3n-1-2n^3+1$

$=-5n^2-5n=-5n\left(n+1\right)⋮5$

Vậy $\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1⋮5$

Bài 2: Nhân đa thức với đa thức