Bài 1:
\(a.A=n^2\left(n+1\right)+2n\left(n+1\right)\\ =\left(n+1\right)\left(n^2+2n\right)\\ =n\left(n+1\right)\left(n+2\right)\)
n(n+1)(n+2) là tích của 3 số tự nhiên liên tiếp nên chia hết cho 2 và 3
Mà: (2;3) = 1 => A chia hết cho 2*3 = 6
\(b.B=\left(4n+3\right)^2-25=\left(4n+3\right)^2-5^2\\ =\left(4n+3-5\right)\left(4n+3+5\right)\\ =\left(4n-2\right)\left(4n+8\right)=8\left(2n-1\right)\left(n+2\right)⋮8\)
\(c.C=\dfrac{n}{3}+\dfrac{n^2}{2}+\dfrac{n^3}{6}\\ =\dfrac{3n^2+2n+n^3}{6}\\ =\dfrac{n\left(n^2+3n+2\right)}{6}=\dfrac{n\left(n+1\right)\left(n+2\right)}{6}\)
n(n + 1)(n + 2) là tích của 3 số liên tiếp => chia hết cho 6
=> C nguyên
Bài 2:
a: \(A=x^4+2x^3+4x^2+8x\)
\(=x^3\left(x+2\right)+4x\left(x+2\right)=\left(x+2\right)\left(x^3+4x\right)\)
\(=x\left(x+2\right)\left(x^2+4\right)\)
Khi x=3 thì \(A=3\left(3+2\right)\left(3^2+4\right)=3\cdot5\cdot13=195\)
b: \(B=x^7-x^6+x^5-x^4+x^3-x^2+x-1\)
\(=x^6\left(x-1\right)+x^4\left(x-1\right)+x^2\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^6+x^4+x^2+1\right)\)
\(=\left(x-1\right)\left(x^2+1\right)\left(x^4+1\right)\)
Khi x=2 thì \(B=\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)=5\cdot17=85\)
c: \(C=x^6-x^5+x^4-x^3+x^2-x\)
\(=x^5\left(x-1\right)+x^3\left(x-1\right)+x\left(x-1\right)=\left(x-1\right)\left(x^5+x^3+x\right)\)
\(=x\left(x^4+x^2+1\right)\left(x-1\right)\)
Khi x=2 thì \(C=2\left(2^4+2^2+1\right)\left(2-1\right)=2\cdot\left(16+4+1\right)=2\cdot21=42\)
d: \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)
\(=20x^3-10x^2-4x-20x^3+10x^2+5x=x\)
Khi x=-5 thì D=x=-5
