(1+y-x^2+x^2y-2xy)(1+y-x^2+x^2y+2xy)分解因式

jianghunima2022-10-04 11:39:547条回答

(1+y-x^2+x^2y-2xy)(1+y-x^2+x^2y+2xy)分解因式
(1+y-x^2+x^2y-2xy)(1+y-x^2+x^2y+2xy)
最后要化成
(x-1)(xy-x-1-y)(x+1)(xy-x+1+y)
说出道理
2l
好象复杂了些..............
..............
其实应该很简单...........
1l
什么乱器八糟的..........不就是想挣分嘛........
好好答..........

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单行道逆驶 共回答了25个问题 | 采纳率88%
(1+y-x^2+x^2y-2xy)(1+y-x^2+x^2y+2xy)分解因式
=[(1-x^2)+(x^2y-2xy+y)][(1-x^2)+(x^2y+2xy+y)]
=[(1-x^2)+y(x^2-2x+1)][(1-x^2)+y(x^2+2x+1)]
=[(1+x)(1-x)+y(x-1)^2][(1+x)(1-x)+y(x+1)^2]
=(x-1)[-(x+1)+y(x-1)](x+1)[1-x+y(x+1)]
=(x-1)(xy-x-y-1)(x+1)(xy-x+y+1)
1年前
地方批 共回答了8个问题 | 采纳率
楼上的好厉害
我算了半天也没弄明白哎
1年前
45xue 共回答了7个问题 | 采纳率
=(x-1)(xy-x-y-1)(x+1)(xy-x+y+1)
1年前
xiaoyanyan_2005 共回答了1个问题 | 采纳率
1+y-x^2+x^2y-2xy)(1+y-x^2+x^2y+2xy)分解因式
=[(1-x^2)+(x^2y-2xy+y)][(1-x^2)+(x^2y+2xy+y)]
=[(1-x^2)+y(x^2-2x+1)][(1-x^2)+y(x^2+2x+1)]
=[(1+x)(1-x)+y(x-1)^2][(1+x)(1-x)+y(x+1)^2]
=(x-1)[-(x+1)+y(x-1)](x+1)[1-x+y(x+1)]
=(x-1)(xy-x-y-1)(x+1)(xy-x+y+1)
1年前
zealkenan 共回答了5个问题 | 采纳率
1
1年前
mavisliang 共回答了1个问题 | 采纳率
哗~
真强~
自愧不如
厉害
1年前
prudent 共回答了11个问题 | 采纳率
(1+y-x^2+x^2y-2xy)(1+y-x^2+x^2y+2xy)分解因式
=[(1-x^2)+(x^2y-2xy+y)][(1-x^2)+(x^2y+2xy+y)]
=[(1-x^2)+y(x^2-2x+1)][(1-x^2)+y(x^2+2x+1)]
=[(1+x)(1-x)+y(x-1)^2][(1+x)(1-x)+y(x+1)^2]
=(x-1)[-(x+1)+y(x-1)](x+1)[1-x+y(x+1)]
=(x-1)(xy-x-y-1)(x+1)(xy-x+y+1)不错不错我也是这样想DI
1年前

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happy路 共回答了12个问题 | 采纳率91.7%
原式=xy(x-y)+(x-y)(x^2+xy+y^2)=(x-y)(x^2+2xy+y^2)=(x-y)(x+y)^2