a )   A = - 2 y 3   –   4 .                                       b )   B =   5 x n .

=(a+b+c)(a2+b2+c2−ab−bc−ca)

=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)

=(a+b)3+c3−3ab(a+b+c)

=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b

a3+b3+c3−3abc

Ta có N = 2 x n ( 3 x n + 2   –   1 )   –   3 x n + 2 ( 2 x n   –   1 )

N = 2 x n ( 3 x n + 2   –   1 )   –   3 x n + 2 ( 2 x n   –   1 )

= 2 x n .3 x n + 2 − 2 x n .1 − 3 x n + 2 .2 x n − 3 x n + 2 . − 1

=   6 x n + n + 2   –   2 x n   –   6 . x n + 2 + n   +   3 x n + 2     =   6 x 2 n + 2   –   6 x 2 n + 2   –   2 x n   +   3 x n + 2     =   –   2 x n   +   3 x n + 2

Vậy N = –   2 x n   +   3 x n + 2

Đáp án cần chọn là: C

\(=-x^2y^3\cdot2x^{n-2}y^n+x^2y^3\cdot3x^ny^{n-3}-x^2y^3\cdot x^{n-2}y^{n-3}\)

\(=-2x^ny^{n+3}+3x^{n+2}y^n-x^ny^n\)

n = 6

\(=3^{2n}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n-2}-y^{2n}\)

\(=3x^{2n}-y^{2n}\)

\(\dfrac{3x^{n+1}y^2-2x^5y^n+x^4y^2}{2x^4y^{n-2}}=\dfrac{3}{4}x^{n+1-4}\cdot y^{2-n+2}-x^{5-4}\cdot y^{n-n+2}+\dfrac{1}{2}x^{4-4}\cdot y^{2-n+2}\)

\(=\dfrac{3}{4}x^{n-3}y^{4-n}-xy^2+\dfrac{1}{2}y^{4-n}\)

Để đây là phép chia hết thì n-3>=0 và 4-n>=0

=>3<=n<=4

=>n=3;n=4

\(5\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+2y^{n-1}\right)+4\left(-x^{n+1}+2y^{n-1}\right)\)

\(=15x^{n+1}-5y^{n-1}-3x^{n+1}-6y^{n-1}-4x^{n+1}+8y^{n-1}\)

\(=8x^{n+1}-3y^{n-1}\)