f(x)=(4^x+4^(-x)-2)/(4^x+4^(-x)+2)
=[(2^x)²-2+(2^-x)²]/[(2^x)²+2+(2^-x)²]
=(2^x-2^-x)²/(2^x+2^x)²
所以√f(x)=(2^x-2^-x)/(2^x+2^-x)
√f(y)=(2^y-2^-y)/(2^y+2^-y)
故(√f(x)+√f(y))/1+√f(x)*f(y)
=[(2^x-2^-x)/(2^x+2^-x)+(2^y-2^-y)/(2^y+2^-y)]/[1-(2^x-2^-x)/(2^x+2^-x)(2^y-2^-y)/(2^y+2^-y)]
=[(2^x-2^-x)(2^y+2^-y)+(2^y-2^-y)(2^x+2^-x)]/[(2^x+2^-x)(2^y+2^-y)+(2^x-2^-x)(2^y-2^-y)]
=[2^(x+y)-2^(-x-y)]/[2^(x+y)+2^(-x-y)]
=√f(x+y)
得证
f(x)=(4^x+4^(-x)-2)/(4^x+4^(-x)+2)=(2^x-2^(-x))^2/(2^x+2^(-x))^2
当x>0时,2^x>1>2^(-x)
所以√f(x)=(2^x-2^(-x))/(2^x+2^(-x))
所以左边=√f(x+y)=(2^(x+y)-2^(-x-y))/(2^(x+y)+2^(-x-y))
右边=(√f(x)+√f(y)/(1+√f(x)√f(y))
=((2^x-2^(-x))(2^y+2^(-y))+(2^x+2^(-x))(2^y-2^(-y)))/
((2^x+2^(-x))(2^y+2^(-y))+(2^x-2^(-x))(2^y-2^(-y)))
=(2*2^(x+y)-2*2^(-x-y))/(2*2^(x+y)+2*2^(-x-y))
=(2^(x+y)-2^(-x-y))/(2^(x+y)+2^(-x-y))=左边
所以原等式成立:√f(x+y)=(√f(x)+√f(y))/(1+√(f(x)*f(y)))
x>0,y>0
好难