已知:x+y=1,xy=-½,利用因式分解求x(x^2-y^2)-x(x+y)^2的值
答案:6 悬赏:80
解决时间 2021-02-07 17:48
- 提问者网友:霸道又专情♚
- 2021-02-06 21:19
已知:x+y=1,xy=-½,利用因式分解求x(x^2-y^2)-x(x+y)^2的值
最佳答案
- 二级知识专家网友:努力只為明天
- 2021-02-06 21:24
x(x^2-y^2)-x(x+y)^2
=x(x-y)(x+y)-x(x+y)^2
=x(x+y)[(x-y)-(x+y)]
=-2xy(x+y)
=1
=x(x-y)(x+y)-x(x+y)^2
=x(x+y)[(x-y)-(x+y)]
=-2xy(x+y)
=1
全部回答
- 1楼网友:一场云烟
- 2021-02-07 03:09
后一个是不是y(x+y)^2
- 2楼网友:劳资的心禁止访问
- 2021-02-07 02:02
x(x+y)(x-y)-x(x+y)^2的公因式为x(x+y),提公因式得:
原式=x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=-2xy(x+y),
代入已知条件x+y=1,xy=-1/2得
原式=(-2)*(-1/2)=1
- 3楼网友:放肆的依賴
- 2021-02-07 00:22
x^2-y^2=(x+y)(x-y)
x(x^2-y^2)-x(x+y)^2=x(x+y)(x-y-x-y)= -2xy(x+y)=1
- 4楼网友:ー何必说爱
- 2021-02-06 23:10
x(x^2-y^2)-x(x+y)^2
=x(x+y)(x-y)-x(x+y)^2
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(-2y)
=-2xy(x+y)
当x+y=1,xy=-½时
原式=1X1=1
- 5楼网友:情战凌云蔡小葵
- 2021-02-06 22:24
x(x^2-y^2)-x(x+y)^2
=x[(x^2-y^2)-(x+y)^2]
=x[x^2-y^2-(x^2+y^2+2xy)]
=x(x^2-y^2-x^2-y^2-2xy)
=-2xy(x+y)
=1
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