Question

已知实数mn满足m2+m-2009=0,1/n2-1/n-2009=0则1/m+n=

Answer 1

由m2+m-2009=0,1/n2-1/n-2009=0

显然方程x^2+x-2009=0有两根m,-1/n.根据韦达定理得

m-1/n=-1 m/n=2009.即1/n=m+1,n=1/(1+m)

又1/m+n=(1+mn)/m=(1+m/(1+m))/m=1/m+1/(1+m)=1/(m^2+m)=1/2009.

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