Question

初二级数学整式的乘除

1.(2a-b)^2*(b+2a)^2 2.(a-1)*(a^4+1)*(a^2+1)*(a+1) 3.(2x+3y)(4x+5y)(2x-3y)(5y-4x)
每步都要算出来

Answer 1

(1)解原式=(2a+b)^2*(2a-b)^2
=(4a^2-b^2)^2
=16*a^4-8*a^2*b^2+b^4
(2)解原式=(a-1)(a^4+1)(a^2+1)(a+1)
=(a^4+1)(a^2+1)(a^2-1)
=(a^4+1)(a^4-1)
=a^8-1
(3)解原式=(2x+3y)(2x-3y)(5y-4x)(5y+4x)
=(4x^2-9y^2)(25y^2-16x^2)

Answer 2

身边没带纸和笔

Answer 3

1.原式=(2a+b)^2*(2a-b)^2=(4a^-b^)^2=16*a^4-8*a^2*b^2+b^4结束.2.原式=(a-1)(a^4+1)(a^2+1)(a+1)=(a^4+1)(a^2+1)(a^2-1)=(a^4+1)(a^4-1)=a^8-1结束.3.原式=(2x+3y)(2x-3y)(5y-4x)(5y+4x)=(4x^2-9y^2)(25y^2-16x^2)