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设非零复数x,y满足x^2+xy+y^2=0,求代数式(x/x+y)^1990+(y/x+y)^1990的值
人气:266 ℃ 时间:2020-04-08 16:11:52
解答
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
同样(y/x)^3=1`
所以[x/(x+y)]^1990+[y/(x+y)]^1990
=(y/x)^1990+(x/y)^1990
=(y/x)^2+(x/y)^2
=-1-y/x-1-x/y
=-2-(x^2+y^2)/xy
=-1
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