(x+y)^4+(x^2-y^2)^2+(x-y)^4 因式分解
答案:2 悬赏:70 手机版
解决时间 2021-03-07 16:15
- 提问者网友：niaiwoma
- 2021-03-07 08:41
用“换元法”解出
最佳答案
- 五星知识达人网友：怙棘
- 2021-03-07 09:16
(x+y)^4+(x^2-y^2)^2+(x-y)^4
=(x+y)^4+(x+y)^2(x-y)^2+(x-y)^4
=(x+y)^4+2(x+y)^2(x-y)^2+(x-y)^4-(x+y)^2(x-y)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=(2x^2+2y^2)^2-(x^2-y^2)^2
=[(2x^2+2y^2)+(x^2-y^2)][(2x^2+2y^2)-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)
用换元法的话，就把x+y=a.x-y=b，代入，最后回代即可
=(x+y)^4+(x+y)^2(x-y)^2+(x-y)^4
=(x+y)^4+2(x+y)^2(x-y)^2+(x-y)^4-(x+y)^2(x-y)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=(2x^2+2y^2)^2-(x^2-y^2)^2
=[(2x^2+2y^2)+(x^2-y^2)][(2x^2+2y^2)-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)
用换元法的话，就把x+y=a.x-y=b，代入，最后回代即可
全部回答
- 1楼网友：动情书生
- 2021-03-07 10:46
(x+y)^4+(x^2-y^2)^2+(x-y)^4
=((x+y)^2)^2+((x-y)^2)^2+2(x+y)^2(x-y)^2-2(x+y)^2(x-y)^2+(x^2-y^2)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=(2x^2+2y^2)^2-(x^2-y^2)^2
=(2x^2+2y^2+x^2-y^2)(2x^2+2y^2-x^2+y^2)
=(3x^2+y^2)(x^2+3y^2)
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