已知f(x)=e^x-e^-x,g(x)=e^x+e^-x,设f(x)f(y)=4,g(x)(y)=8,求[g（x+y）

f(-x)=-f(x)
g(-x)=g(x)
g(x+y)
=e^(x+y)+e^-(x+y)
=1/2*[(e^x+e^-x)(e^y+e^-y) + (e^x-e^-x)(e^y-e^-y)]
=1/2*[g(x)g(y)+f(x)f(y)]
g(x-y) 【-y 代替y】
=1/2*[g(x)g(y)-f(x)f(y)]
[g（x+y）]/[g（x-y）]
={1/2*[g(x)g(y)+f(x)f(y)]}/{1/2*[g(x)g(y)-f(x)f(y)]}
= 3