f(x)=e^x-e^-x,g(x)=e^x+e^-x,e=2.71(1)若f(x)f(y)=4,g(x)g(y)=8,
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f(x)=e^x-e^-x,g(x)=e^x+e^-x,e=2.71(1)若f(x)f(y)=4,g(x)g(y)=8,求g(x+y)/g(x-y)?
f(x)f(y)=[e^x-e^-x][e^y-e^-y]
=e^(x+y)-e^(x-y)-e^[-(x-y)]+e^[-(x+y)]
=g(x+y)-g(x-y)=4
g(x)g(y)=[e^x+e^-x][e^y+e^-y]
=e^(x+y)+e^(x-y)+e^[-(x-y)]+e^[-(x+y)]
=g(x+y)+g(x-y)=8
解得:
g(x+y)=6
g(x-y)=2
因此:
g(x+y)/g(x-y)=3
=e^(x+y)-e^(x-y)-e^[-(x-y)]+e^[-(x+y)]
=g(x+y)-g(x-y)=4
g(x)g(y)=[e^x+e^-x][e^y+e^-y]
=e^(x+y)+e^(x-y)+e^[-(x-y)]+e^[-(x+y)]
=g(x+y)+g(x-y)=8
解得:
g(x+y)=6
g(x-y)=2
因此:
g(x+y)/g(x-y)=3
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