1个回答
展开全部
x^4+7x^3+14x^2+7x+1
= x^4+7x^3+12x^2 + 2x^2+7x + 1
= (x^2+3x)(x^2+4x) + x^2+3x + x^2+4x + 1
=(x^2+3x)(x^2+4x+1)+(x^2+4x+1)
= (x^2+3x+1)(x^2+4x+1)
(x+y)^2-1=(x+y+1)(x+y-1)
所以原式
=(x+y+1)(x+y-1)-2xy(x+y-1) 这个式子把(x+y-1)提出来
=(x-2xy+y+1)(x+y-1)
= x^4+7x^3+12x^2 + 2x^2+7x + 1
= (x^2+3x)(x^2+4x) + x^2+3x + x^2+4x + 1
=(x^2+3x)(x^2+4x+1)+(x^2+4x+1)
= (x^2+3x+1)(x^2+4x+1)
(x+y)^2-1=(x+y+1)(x+y-1)
所以原式
=(x+y+1)(x+y-1)-2xy(x+y-1) 这个式子把(x+y-1)提出来
=(x-2xy+y+1)(x+y-1)
追问
第二题我写错了,是 (x+y)^3+2xy(1-x-y)-1
追答
(x+y)^3+2xy(1-x-y)-1
=(x+y-1)((x+y)^2+2(x+y)+1)-2xy(x+y-1)
=(x+y-1)(x^2+y^2+2xy+2x+2y+1-2xy)
=(x+y-1)(x^2+y^2+2x+2y+1)
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