已知x+y=1，xy=-二分之一，求x（x+y)(x-y)-x(x+y)²
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解决时间 2021-04-06 05:01
- 提问者网友：不爱我么
- 2021-04-05 05:12
已知x+y=1，xy=-二分之一，求x（x+y)(x-y)-x(x+y)²
最佳答案
- 五星知识达人网友：山君与见山
- 2021-04-05 05:26
x（x+y)(x-y)-x(x+y)²
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(x-y-x-y)
=-2xy(x+y)
=-2×(-1/2)×1
=1
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(x-y-x-y)
=-2xy(x+y)
=-2×(-1/2)×1
=1
全部回答
- 1楼网友：旧脸谱
- 2021-04-05 06:11
X(X^2-Y^2)-X(X^2+Y^2+2XY)=x^3-xy^2-x^3+xY^2-2xy=2xy=-1
- 2楼网友：鸽屿
- 2021-04-05 05:43
解：由已知，有
x+y=1，xy=-1/2，x²-x=1/2
于是
x（x+y)(x-y)-x(x+y)²
=x·1·(x-y)-x·1²
=x²-xy-x
=x²-（-1/2）-x
=x²-x+1/2
=1/2+1/2
=1
或者先化简：
x（x+y)(x-y)-x(x+y)²
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(x-y-x-y)
=-2xy(x+y)
=-2×(-1/2)×1
=1
x+y=1，xy=-1/2，x²-x=1/2
于是
x（x+y)(x-y)-x(x+y)²
=x·1·(x-y)-x·1²
=x²-xy-x
=x²-（-1/2）-x
=x²-x+1/2
=1/2+1/2
=1
或者先化简：
x（x+y)(x-y)-x(x+y)²
=x(x+y)[(x-y)-(x+y)]
=x(x+y)(x-y-x-y)
=-2xy(x+y)
=-2×(-1/2)×1
=1
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