f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f`(ξ)/1-ξ.f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证:存在ξ∈(0,1),使f``(ξ)=2f(ξ)/1-ξ.
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