若非零复数x,y满足x²+xy+y²=0,则【x/(x+y)】^2005+【y/(x+y)】^200
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若非零复数x,y满足x²+xy+y²=0,则【x/(x+y)】^2005+【y/(x+y)】^2005的值是?
为了问答双方利益抄题目已校对,
为了问答双方利益抄题目已校对,
x^2+xy+y^2=0,两边除以y^2得,(x/y)^2+x/y+1=0,
根据求根公式得,x/y=(-1+√3i)/2,或x/y=(-1-√3i)/2,
当x/y=(-1+√3i)/2时,x/(x+y)=1-y/(x+y)=1-1/(x/y+1)=(1+√3i)/2,
y/(x+y)=(1-√3i)/2,
当x/y=(-1-√3i)/2时,x/(x+y)=(1-√3i)/2,y/(x+y)=(1+√3i)/2,
两种情况都是[x/(x+y)]^2005+[y/(x+y)]^2005=[(1-√3i)/2]^2005+[(1+√3i)/2]^2005,
计算[(1-√3i)/2]^3=-1,[(1+√3i)/2]^3=-1,
2005/3=668...1
所以原式=(-1)^1+(-1)^1=-2
参考以下做法:
x^2=-y(x+y)
y^2=-x(x+y)
所以x/(x+y)=-y/x
y/(x+y)=-x/y
而(x/y)^2+x/y+1=0
所以(x/y)^3-1=(x/y-1)((x/y)^2+x/y+1)=0
(x/y)^3=1
同样(y/x)^3=1`
所以[x/(x+y)]^2005+[y/(x+y)]^2005
=(-y/x)^2005+(-x/y)^2005
=(-y/x)^1+(-x/y)^1
=-1-1
=-2
根据求根公式得,x/y=(-1+√3i)/2,或x/y=(-1-√3i)/2,
当x/y=(-1+√3i)/2时,x/(x+y)=1-y/(x+y)=1-1/(x/y+1)=(1+√3i)/2,
y/(x+y)=(1-√3i)/2,
当x/y=(-1-√3i)/2时,x/(x+y)=(1-√3i)/2,y/(x+y)=(1+√3i)/2,
两种情况都是[x/(x+y)]^2005+[y/(x+y)]^2005=[(1-√3i)/2]^2005+[(1+√3i)/2]^2005,
计算[(1-√3i)/2]^3=-1,[(1+√3i)/2]^3=-1,
2005/3=668...1
所以原式=(-1)^1+(-1)^1=-2
参考以下做法:
x^2=-y(x+y)
y^2=-x(x+y)
所以x/(x+y)=-y/x
y/(x+y)=-x/y
而(x/y)^2+x/y+1=0
所以(x/y)^3-1=(x/y-1)((x/y)^2+x/y+1)=0
(x/y)^3=1
同样(y/x)^3=1`
所以[x/(x+y)]^2005+[y/(x+y)]^2005
=(-y/x)^2005+(-x/y)^2005
=(-y/x)^1+(-x/y)^1
=-1-1
=-2
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