(x+2y)的平方乘以(x的平方+4y的平方)的平方乘以(x-2y)的平方-(x的四次方-2y的四次方)的平方

简便算法,最好给我讲一下
2025-03-28 10:52:02
推荐回答(2个)
回答1:

=[(x-2y)(x+2y)]^2(x^2+4y^2)^2-[x^4-(2y)^4]^2
=(x^2-4y^2)^2(x^2+4y^2)^2-[x^4-(2y)^4]^2
=(x^4-16y^4)^2-[x^4-16y^4]^2
=0

回答2:

[(x+2y)^2*(x-2y)^2]^2*(x^2+4y^2)^2-[x^4-(2y)^4]^2
=[(x+2y)(x-2y)]^2*(x^2+4y^2)^2-[x^4-(2y)^4]^2
=(x^2-4y^2)^2*(x^2+4y^2)^2-[x^4-(2y)^4]^2
=[(x^2-4y^2)(x^2+4y^2)]^2-[x^4-(2y)^4]^2
=(x^4-16y^4)^2-[x^4-(2y)^4]^2
=0