y=(1/2)^(x^2-6x+10)的定义域是实数集Ry=(1/2)^(x^2-6x+10)=2^(-(x-3)^2-1)f(x)=-(x-3)^2-1在实数集上取x=3时有最大值所以y=(1/2)^(x^2-6x+10)的最大值为2^(-1)=1/2而函数y=(1/2)^(x^2-6x+10)恒大于零,所以值域为（0...