Question

x4+(x+y)4+y4 分解因式

最好有步骤，谢谢啦~~

Answer 1

分析 这是一个二元对称式,二元对称式的基本对称式是x+y,xy任何二元对称多项式都可用x+y,xy表示,如x2+y2=(x+y)2-2xy,二元对称多项式的分解方法之一是:先将其用xy,x+y表示,再行分解.

解 ∵x4+y4

=(x+y)4-4x3y-6x2y2-4xy2

=(x+y)4-4xy(x+y)2+2x2y2.

∴原式=(x+y)4-4xy(x+y)2+2x2y2+(x+y)4

=2(x+y)4-4xy(x+y)2+2x2y2

=2[(x+y)4-2xy(x+y)2+(xy)2]

=2[(x+y)2-xy]2-2(x2+y2+xy)2,

Answer 2

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

Answer 3

解答:解:x4+y4+(x+y)4-2，
=(x2+y2)2-2x2y2+(x2+2xy+y2)2-2，
=(x2+y2)2-2x2y2+(x2+y2)2+4xy(x2+y2)+4x2y2-2，
=2(x2+y2)2+2x2y2+4xy(x2+y2)-2，
=2[(x2+y2)2+x2y2+2xy(x2+y2)-1]，
=2[(x2+xy+y2)2-1]，
=2(x2+xy+y2-1)(x2+xy+y2+1).
故答案为:2(x2+xy+y2-1)(x2+xy+y2+1).

Answer 4

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

Answer 5

x4+y4+(x+y)4-2=(x2+y2)2-2x2y2+(x+y)4-2
=[(x+y)2-2xy]2-2x2y2+(x+y)4-2
=(x+y)4-4xy(x+y)2+4x2y2-2x2y2+(x+y)4-2
=2(x+y)4-4xy(x+y)2+2x2y2-2
=2[(x+y)4-2xy(x+y)2+x2y2]-2
=2[(x+y)2-xy]2-2
=2[x2+y2+xy]2-2=2(x2+y2+xy+1)(x2+y2+xy-1)