题目
（x+y）^2-z^=
a^2-(b-c)^2=
(x-y)^3-x+y=
16(x+y)^2-25(x-y)^2=
a^2-(b-c)^2=
(x-y)^3-x+y=
16(x+y)^2-25(x-y)^2=
提问时间：2020-11-03
答案
（x+y）^2-z^2
=(x+y+z)(x+y-z)
a^2-(b-c)^2
=(a-b+c)(a+b-c)
(x-y)^3-x+y
=(x-y)^3-(x-y)
=(x-y)[(x-y)^2-1]
=(x-y)(x-y+1)(x-y-1)
16(x+y)^2-25(x-y)^2
=[4(x+y)]^2-[5(x-y)]^2
=[4(x+y)+5(x-y)]*[4(x+y)-5(x-y)]
=(4x+4y+5x-5y)(4x+4y-5x+5y)
=(9x-y)(9y-x)
=(x+y+z)(x+y-z)
a^2-(b-c)^2
=(a-b+c)(a+b-c)
(x-y)^3-x+y
=(x-y)^3-(x-y)
=(x-y)[(x-y)^2-1]
=(x-y)(x-y+1)(x-y-1)
16(x+y)^2-25(x-y)^2
=[4(x+y)]^2-[5(x-y)]^2
=[4(x+y)+5(x-y)]*[4(x+y)-5(x-y)]
=(4x+4y+5x-5y)(4x+4y-5x+5y)
=(9x-y)(9y-x)
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