分解因式(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd
答案:4 悬赏:60 手机版
解决时间 2021-07-24 17:42
- 提问者网友:我的未来我做主
- 2021-07-24 07:08
分解因式(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd
最佳答案
- 五星知识达人网友:轻雾山林
- 2021-07-24 08:48
(注:^代表平方,打不出来啊)
解:原式=[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd
=[(a+b)^-(c+d)^][(c-d)^-(b-a)^]+16abcd
=(a^+b^-c^-d^+2ab-2cd)(c^+d^-a^-b^-2cd+2ab)+16abcd
=(2ab-2cd)^-(a^+b^-c^-d^)^+16abcd
=4a^b^+4c^d^-8abcd+16abcd-(a^+b^-c^-d^)
=4a^b^+4c^d^+8abcd-(a^+b^-c^-d^)
=(2ab+2cd)^-(a^+b^-c^-d^)^
=(2ab+2cd+a^+b^-c^-d^)(2ab+2cd-a^-b^+c^+d^)
=......
=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)
(思路:前面4个等号都是用平方差公式,第5等号解决掉+16abcd,之后基本是照原路返回)^_^
解:原式=[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd
=[(a+b)^-(c+d)^][(c-d)^-(b-a)^]+16abcd
=(a^+b^-c^-d^+2ab-2cd)(c^+d^-a^-b^-2cd+2ab)+16abcd
=(2ab-2cd)^-(a^+b^-c^-d^)^+16abcd
=4a^b^+4c^d^-8abcd+16abcd-(a^+b^-c^-d^)
=4a^b^+4c^d^+8abcd-(a^+b^-c^-d^)
=(2ab+2cd)^-(a^+b^-c^-d^)^
=(2ab+2cd+a^+b^-c^-d^)(2ab+2cd-a^-b^+c^+d^)
=......
=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)
(思路:前面4个等号都是用平方差公式,第5等号解决掉+16abcd,之后基本是照原路返回)^_^
全部回答
- 1楼网友:一秋
- 2021-07-24 11:35
[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd
=[(a+b)^2-(c+d)^2][(c-d)^2-(b-a)^2]+16abcd
=(a^2+b^2-c^2-d^2+2ab-2cd)(c^2+d^2-a^2-b^2-2cd+2ab)+16abcd
=(2ab-2cd)^2-(a^2+b^2-c^2-d^2)^2+16abcd
=4a^2b^2+4c^2d^2-8abcd+16abcd-(a^2+b^2-c^2-d^2)^2
=4a^2b^2+4c^2d^2+8abcd-(a^2+b^2-c^2-d^2)^2
=(2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2
=(2ab+2cd+a^2+b^2-c^2-d^2)(2ab+2cd-a^2-b^2+c^2+d^2)
=[(a+b)^2-(c-d)^2][(c+d)^2-(a-b)^2]
=(a+b+c-d)(a+b-c+d)(a-b+c+d)(c+d-a+b)
- 2楼网友:舍身薄凉客
- 2021-07-24 10:55
(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd
=[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd
=[(a+b)²-(c+d)²][(c-d)²-(b-a)²]+16abcd
=(a²+b²-c²-d²+2ab-2cd)(c²+d²-a²-b²-2cd+2ab)+16abcd
=(2ab-2cd)²-(a²+b²-c²-d²)²+16abcd
=4a²b²+4c²d²-8abcd+16abcd-(a²+b²-c²-d²)
=4a²b²+4c²d²+8abcd-(a²+b²-c²-d²)
=(2ab+2cd)²-(a²+b²-c²-d²)²
=(2ab+2cd+a²+b²-c²-d²)(2ab+2cd-a²-b²+c²+d²)
=[(a+b)²-(c-d)²][(c+d)²-(a-b)²]
=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)
- 3楼网友:撞了怀
- 2021-07-24 09:16
(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd
=[(b+c)+(a+d)][(b+c)-(a+d)][(a-d)+(c-b)][(a-d)-(c-b)]+16abcd
=[(b+c)^2-(a+d)^2][(a-d)^2-(c-b)^2]+16abcd
=
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息,可以点下面链接进行举报!
大家都在看
推荐资讯