设函数f(x)=2的X次方+2的X次方分之a再 —1(a为常数)

(1) a=0时,f(x) = 2^x - 1x>0时,g(x)=f(x) = 2^x - 1x0f(x) = t + a/t - 1 = 0 t² - t + a = 0t = [1±√(1-4a) ] / 2[1-√(1-4a) ] / 2 < 0 舍去t= 1+√(1-4a) = 2^xx = log₂ [ 1+√(1-4a) ]是的 不过最后答案错了更正：t= [1+√(1-4a)] /2 = 2^xx = log₂ [ 1+√(1-4a) ] - 1