1) \(x^3+6x^2+11x+6\)

\(=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+5x+6\right)\)

\(=\left(x+1\right)\left(x^2+2x+3x+6\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

2) \(A=n^3\left(n^2-7\right)^2-36n\)

\(A=n\left[n^2\left(n^2-7\right)^2-36\right]\)

\(A=n\left\{\left[n\left(n^2-7\right)\right]^2-6^2\right\}\)

\(A=n\left(n^3-7n-6\right)\left(n^3-7n+6\right)\)

\(A=n\left(n^3-7n-6\right)\left(n^3-n-6n+6\right)\)

\(A=n\left(n^3-7n-6\right)\left[n\left(n-1\right)\left(n+1\right)-6\left(n-1\right)\right]\)

\(A=n\left(n^3-7n-6\right)\left(n-1\right)\left(n^2+n-6\right)\)

\(A=n\left(n-1\right)\left(n^3-7n-6\right)\left(n^2+3n-2n-6\right)\)

\(A=n\left(n-1\right)\left(n^3-7n-6\right)\left[n\left(n+3\right)-2\left(n+3\right)\right]\)

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n^3-7n-6\right)\)

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

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left[n\left(n-1\right)\left(n+1\right)-6\left(n+1\right)\right]\)

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

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

\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left[n\left(n+3\right)-2\left(n+3\right)\right]\)

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

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

Rồi sao nữa còn nghĩ :))

Bạn đã giải được bài này chưa?

B = n³(n²-7)^2-36n
   = n³(n⁴-14n²+49)-36n
   = n⁷ - 14n⁵ + 49n³ - 36n
   = n(n⁶ - 14n⁴ +49n² -36)
   = n(n⁶ - n⁵ + n⁵ - n⁴ - 13n⁴ + 13n³ - 13n³ + 13n² + 36n² - 36n + 36n - 36)
   = n[n⁵(n-1)+n⁴(n-1)-13n³(n-1)-13n²(n-1)+36n(n-1)+36(n-1)]
   = n(n-1)(n⁵+n⁴-13n³-13n²+36n+36)
   = n(n-1)[n⁴(n+1)-13n²(n+1)+36(n+1)]
   = n(n-1)(n+1)(n⁴-13n²+36)
   = n(n-1)(n+1)(n⁴-9n²-4n²+36)
   = n(n-1)(n+1)[n²(n²-9)-4(n²-9)]
   = n(n-1)(n+1)(n²-9)(n²-4)
   = n(n-1)(n+1)(n-3)(n+3)(n-2)(n+2)
   = (n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)
Có \(B⋮3\); \(B⋮5\);\(B⋮7\)(vì có 7 số tự nhiên liên tiếp)
Mà 3; 5; 7 đôi một nguyên tố cùng nhau
\(\Rightarrow B⋮3.5.7\Rightarrow B⋮105\)(đpcm)

Phạm Hồ Thành Quang làm đúng đấy

a,\(5n^3+15n^2+10n=5n\left(n^2+3n^2+2\right)=5n\left(n^2+n+2n+2\right)=5n\left(n+1\right)\left(n+2\right)\)Nhận thấy 5n(n+1)(n+2)\(⋮5\) vì \(5⋮5\) (1)

và \(n\left(n+1\right)\left(n+2\right)⋮6\) vì n(n+1)(n+2) là ba số tự nhiên liên tiếp (2)

Từ (1)và(2)\(\Rightarrow5n\left(n+1\right)\left(n+2\right)⋮30\Rightarrowđpcm\)

b, \(n^3\left(n^2-7\right)-36n\)

\(=n\left[\left(n^2\right)\left(n^2-7\right)^2-36\right]\)

\(=n\left[\left(n^3-7n\right)^2-36\right]\)

\(=n\left(n^3-7n-6\right)\left(n^3-7n+6\right)\)

\(=\left(n-3\right)\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\left(n+3\right)⋮3,5,7\Rightarrow⋮105\Rightarrowđpcm\)

Bn Mai Xuân Phong ơi!Câu a, 5x³hay là 5n³ vậy?

1. \(x^3+6x^2+11x\) +6

= \(x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)

= \(\left(x+3\right)\left(x^2+3x+2\right)\)

=(x+3)(x+1)(x+2)

2. Sua \(n^3\left(n^2+7\right)^2-36n\) thanh \(n^3\left(n^2-7\right)^2-36n\)

A= \(n^3\left(n^2-7\right)^2-36n\)

= \(n^7-14n^5+49n^3-36n\)

= (n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)

Day la tich cua 7 so tu nhien lien tiep=> A \(⋮105\)

Ta có : \(n^3\left(n^2-7\right)^2-36n\)

\(=n[\left(n^3-7n\right)^2-36]\)

\(=n\left(n^3-7n-6\right)\left(n^3-7n+6\right)\)

\(=n[\left(n-3\right)\left(n^2+3n+2\right)][\left(n+3\right)\left(n^2-3n+2\right)]\)

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

là tích của 7 số nguyên liên tiếp 

\(\Rightarrow n\left(n-3\right)\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n-1\right)\left(n-2\right)⋮7\)

hay \(n^3\left(n^2-7\right)^2-36n⋮7\forall n\inℤ\)