证明：设g(x)=ln(1+x),g'(x)=1/(1+x),则g'(m)=1/(1+m)∵f(x),g(x)在[0,a]上连续,在(0,a)内可导,且g'(x)≠0∴由柯西中值定理得至少存在一点m属于(0,a)使得[f(a)-f(0)]/[g(a)-g(0)]=f'(m)/g'(m)即f(a)=(1+m)f'(m)ln(1+...