已知N/(v+u)=N1/v,N/(v-u)=N2/v,怎样解出N=2N1*N2/N1+N2

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已知N/(v+u)=N1/v,N/(v-u)=N2/v,怎样解出N=2N1*N2/N1+N2

已知N/(v+u)=N1/v,N/(v-u)=N2/v,怎样解出N=2N1*N2/N1+N2
已知N/(v+u)=N1/v,N/(v-u)=N2/v,怎样解出N=2N1*N2/N1+N2

已知N/(v+u)=N1/v,N/(v-u)=N2/v,怎样解出N=2N1*N2/N1+N2
分数中有个等比定理,你可以百度下:若a:b=c:d(其中b,d≠0),则(a+c):(b+d)=(a-c):(b-d)=a:b
那么N/(v+u)=N1/v,就可以得到N-N1/(v+u-v)=N1/v,就是左边的分数的分子分母同时减去右边的分子分母,分数值不变,即N-N1/u=N1/v
同理可以得到,N-N2/(-u)=N2/v
然后你把两个等式左边除以左边,右边除以右边,消去u ,v,就得到了