1/m+1/n=4/(m^2+n^2),求m+n的值

南国椰树 |浏览1539次
收藏|2020/01/22 13:06

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2020/01/22 13:38

解法1、1/m + 1/n = 4/(m^2+n^2)解得 m = 1,n = 1,m + n = 2;解法2、2(m^2 + n^2)/2mn = 4/( m + n )由均值不等式,m^2 + n^2≥ 2mn,m = n 时等号成立;故2(m^2 + n^2)/2mn =4/( m + n )≥2,m = n = 1 时,m + n = 2 。

寂园晓月

其他回答(3)
  • 1/m+1/n=4/(m^2+n^2)(m+n)/(mn)=4/(m^2+n^2)m+n=4mn/(m^2+n^2)
    回答于 2020/01/22 14:44
  • 1/m+1/n=4/(m²+n²)设m=kn,k≠01/m+1/n=1/(kn)+1/n=(1+k)/(kn)4/(m²+n²)=4/(k²n²+n²)=4/[n²(k²+1)](k+1)/(kn)=4/[n²(k²+1)]n=4k/[(k+1)(k²+1)]m+n=n(k+1)=4k/(k²+1)1】若k>0m+n=4/(k+1/k)≥4/2=2等号当且仅当k=1时成立。2】若k<0m+n=4/(k+1/k)≤4/(-2)=-2等号当且仅当k=-1时成立。∵4/(m²+n²)>0,∴1/m+1/n>0,∴k<-1或者k>0。∴m+m不是定值。m+n<-2或者m+n≥2。
    回答于 2020/01/22 14:25
  • 1/m+1/n=4/(m²+n²)n/mn+m/mn=4/(m²+n²)(m+n)/mn=4/(m²+n²)m+n=4mn/(m²+n²)
    回答于 2020/01/22 14:00
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