已知f(x)=ex-e-x,g(x)=ex+e-x(e=2.718…).(1)則[f(x)]2-[g(x)]2...

問題詳情：

已知f(x)=ex-e-x,g(x)=ex+e-x(e=2.718…).

(1)則[f(x)]2-[g(x)]2的值為　　　　; 

(2)若f(x)·f(y)=4,g(x)·g(y)=8,則=　　　　. 

【回答】

(1)-4　(2)3

(1)[f(x)]2-[g(x)]2

=[f(x)+g(x)][f(x)-g(x)]

=[(ex-e-x)+(ex+e-x)][(ex-e-x)-(ex+e-x)]

=2ex·(-2e-x)

=-4.

(2)因為f(x)·f(y)=(ex-e-x)(ey-e-y)

=ex+y-ex-y-ey-x+e-(x+y),

g(x)·g(y)=(ex+e-x)(ey+e-y)

=ex+y+ex-y+ey-x+e-(x+y),

g(x+y)=ex+y+e-(x+y),

g(x-y)=ex-y+e-(x-y)=ex-y+ey-x,

所以

解得

所以==3.

*:

知識點：基本初等函式I

題型：填空題