【追加100】三道函数数学题设f(x)=[e^x-e^(-x)]/2,g(x)=[e^x+e^(-x)]/2,求证：(1

设f(x)=[e^x-e^(-x)]/2,g(x)=[e^x+e^(-x)]/2,求证：
(1)[g(x)]^2-[f(x)]^2=1
(2）f(2x)=2f(x)×g(x)
(3)g(2x)=[g(x)]^2+[f(x)]^2
1)
[g(x)]^2-[f(x)]^2=1
[g(x)]^2-[f(x)]^2
=[e^x+e^(-x)]^2/4-[e^x-e^(-x)]^2/4
=[e^2x+e^-2x+2)-(e^2x+e^-2x-2)]/4
=4/4
=1
2)
f(2x)=(e^2x-e^-2x)/2
=(e^x+e^-x)(e^x-e^-x)/2
=2f(x)g(x)
3)
g(2x)=[g(x)]^2+[f(x)]^2
[g(x)]^2+[f(x)]^2
=[e^x+e^(-x)]^2/4+[e^x-e^(-x)]^2/4
=[e^2x+e^-2x+2)+(e^2x+e^-2x-2)]/4
=(2e^2x+2e^-2x)/4
=(e^2x+e^-2x)/2
=g(2x)
得证
已知函数f(x)满足:对于任意实数x,y,f(xy)=f(x)+f(y)
（1）求f(0),f(1),f(-1)的值
（2）求证f(y/x)=f(y)-f(x)
(3)判断函数f(x)奇偶性
2、
(1)
令x=y=0,f(0)=f(0)+f(0),所以f(0)=0
令x=y=1,则f(1)=f(1)+f(1),所以f(1)=0
令x=y=-1,则f(1)=2f(-1)=0.所以f(-1)=0
2)求证f(y/x)=f(y)-f(x)
f(xy)=f(x)+f(y)
f(y)=f(x*y/x)=f(x)+f(y/x)
=>f(y/x)=f(y)-f(x)
3)判断函数f(x)奇偶性
f(-x*-y)=f(-x)+f(-y)=f(xy)
所以函数为偶函数.
已知奇函数f(x)满足f(x+2)=f(-x),而且当x在(0,1)时,f(x)=2的x次方
（1）证明:f(x+4)=f(x)
（2） 求:f(log2^24)的值
1）y=f(x)为奇函数,
故有:
-f(x)=f(-x)=f(x+2)=-f[(x+2)+2]=-f(x+4)
所以f(x)=f(x+4)
得证
2）求:f(log2^24)的值
log2^24=log2^4+log2^2+log2^3
=3+log2^3
f(log2^24)
=f(3+log2^3-4)
=f(log2^3-1)
0

最怕的就是数学！

设f(x)=[e^x-e^(-x)]/2,g(x)=[e^x+e^(-x)]/2,求证：
(1)[g(x)]^2-[f(x)]^2=1
(2）f(2x)=2f(x)×g(x)
(3)g(2x)=[g(x)]^2+[f(x)]^2
1)
[g(x)]^2-[f(x)]^2=1
[g(x)]^2-[f(x)]^2
=[e^x+e^(...

解:
1.(1) 由题可得: [g(x)]^2-[f(x)]^2=[g(x)+f(x)][g(x)-f(x)]
=e^x*e^(-x)=e^0=1;
1.(2) f(2x)=[e^(2x)-e^(-2x)]/2=[(e^x)^2-(e^(-x))^2]/2
=[(e^x+e^(-x))...