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2loga(M-N)=loga(m)+log(N),则N除以M的值
人气:322 ℃ 时间:2020-05-22 11:59:29
解答
2loga(M-N)=loga(m)+log(N),
loga(M-N)^2=loga(M*N)
(M-N)^2=M*N
M^2-2MN+N^2=MN
M^2-3MN+N^2=0
M=(3±√5)/2N
N/M=2/(3±√5)
=2(3±√5)/(6-5)
=2(3±√5)
因为M-N>0 M>0 N>0
M>N
所以0
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