Question

(y+1)平方-2x平方(1+y平方)+x四次方(1-y)平方 高中因式分解~ 急求

Answer 1

(y+1)平方-2x平方(1+y平方)+x四次方(1-y)平方
=(y+1)²+2*x²*(y²-1)+[x²(y-1)]²-2x²(y²+1+y²-1)
=[(y+1)+x(y-1)]²-(2xy)²
=(xy-x+y+1+2xy)(xy-x+y+1-2xy)
=(3xy-x+y+1)(-xy-x+y+1)
=(3xy-x+y+1)[-x(y+1)+(y+1)]
=(3xy-x+y+1)(y+1)(1-x)
希望能帮到你O(∩_∩)O

追问

你好象 错的吧~

追答

是呀，错了
(y+1)平方-2x平方(1+y平方)+x四次方(1-y)平方
=(y+1)²+2x²*(y+1)(y-1)+(x²)²(y-1)²-2x²(y+1)(y-1)-2x²(y²+1)
=(y+1)²+2*x²*(y²-1)+[x²(y-1)]²-2x²(y²-1+y²+1)
=(x²y-x²+y+1)²-(2xy)²
=(x²y+2xy-x²+y+1)(x²y-2xy-x²+y+1)
=[y(x+1)²-(x+1)(x-1)]*[y(x-1)²-(x+1)(x-1)]
=(x+1)(xy+y-x+1)(x-1)(xy-y-x-1)

Answer 2

原式=(1+y^2)+2x^2(1-y^2)+x^4(1-y^2)-2x^2(1-y^2)-2x^2(1+y^2)

　　 =[1+y+x^2(1-y)]^2-2x^2(1-y^2+1+y^2)

=(x^2-x^2y+y+1)^2-4x^2

=(x^2-x^2y+y+1+2x)(x^2-x^2y+y+1-2x)

=[(x^2+2x+1)-y(x^2-1)][(x^2-2x+1)-y(x^2-1)]

=[(x+1)^2-y(x^2-1)][(x-1)^2-y(x^2-1)]

=(x+1)(x+1-xy+y)(x-1)(x-1-xy-y)

追问

你看错题了，平方弄错了

追答

哦哦 sorry啦