Question

6道因式分解求过程

Answer 1

解：（1）(1/27)*a^(3m-n)-(1/3)*a^(m-n)*b^(2n)
=(1/3)*a^(m-n)*[1/9*a^(2m)-b^(2n)]
=(1/3)*a^(m-n)*(1/3*a^m+b^n)(1/3*a^m-b^n)
(2)原式=(2*ab+a^2+b^2)*(2*ab-a^2-b^2)
=(a+b)^2*[-(a-b)^2]=-(a+b)^2*(a-b)^2
(3)原式=x^(n+1)*y^(n-1)*(x^2n+2*x^n*y^n+y^2n)
=x^(n+1)*y^(n-1)*(x^n+y^n)^2
(4)原式=(a^2-1)(b^2-1)[1-(a^2-1)(b^2-1)]
=-(a^2-1)(b^2-1)(a^2*b^2+a^2+b^2)
(5)原式=(x^2+y^2)^2-(y^2+z^2)^2-(z^2-x^2)^2
=(x^2-z^2)(x^2+2*y^2+z^2)-(z^2-x^2)^2
=(x^2-z^2)(x^2+2*y^2+z^2+z^2-x^2)
=2(x^2-z^2)(y^2+z^2)
(6)原式=(a^2-2ab+26*b^2)[25*b^2-(a-b)^2]
=(a^2-2ab+26*b^2)(5b+a-b)(5b-a+b)
=(a^2-2ab+26*b^2)(4b+a)(6b-a)

Answer 2

等我有空了教你