(1) f(x)=(-2^x+b)/(2^x-1)+2
f(-x)=(-2^(-x)+b)/(2^(-x)-1)+2……分子分母同乘以2^x
=(-1+b*2^x)/(1-2^x) +2
函数是奇函数.所以f(-x)= -f(x),
即(-1+b*2^x)/(1-2^x) +2=-(-2^x+b)/(2^x-1)-2,
[(b+1)*2^x-(b+1)]/(2^x-1)=4,
(b+1)*2^x-(b+1)=4*2^x-4,
b+1=4,b=3.
(2)所以 f(x)=(-2^x+3)/(2^x-1)+2……通分得下式
= (2^x +1 )/(2^x – 1)
= (2^x -1+2 )/(2^x – 1)
= 1 + 2/(2^x - 1)
设 x1 < x2
则 f(x2) - f(x1) = 1/(2^x2 - 1) - 1/(2^x1 - 1)
= (2^x1 - 2^x2)/[(2^x2 - 1)(2^x1 - 1)] < 0
所以 f(x2) < f(x1)
所以 f(x)是减函数
(3)f(t²-2t)+f(2 t²-k) k-2 t²
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