因式分解 (x+y)(x+y+2xy)+x的平方y的平方-1
答案:3 悬赏:60
解决时间 2021-02-21 10:26
- 提问者网友:北故人
- 2021-02-21 07:26
因式分解 (x+y)(x+y+2xy)+x的平方y的平方-1
最佳答案
- 二级知识专家网友:安稳不如野
- 2021-02-21 08:13
解:
(x+y)(x+y+2xy)+x的平方y的平方-1
=[(x+y)²+2(x+y)xy+x²y²]-1
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
(x+y)(x+y+2xy)+x的平方y的平方-1
=[(x+y)²+2(x+y)xy+x²y²]-1
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
全部回答
- 1楼网友:為→妳鎖鈊
- 2021-02-21 10:39
原式=(x+y)²+2xy(x+y)+x²y²-1
=(x+y+xy)²-1
=(x+y+xy+1)(x+y+xy-1)
=[x(y+1)+(y+1)](x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
- 2楼网友:最后战士
- 2021-02-21 09:16
令x+y=a,xy=b,则
原式=a(a+2b)+(b+1)(b-1)
=a^2+2ab+b^2-1
=(a+b)^2-1
=(a+b+1)(a+b-1)
故原式=(x+y+xy+1)(x+y+xy-1)
=(x+1)(y+1)(x+y+xy-1)
我要举报
如以上问答内容为低俗、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