18.(2023春•运城期末)已知f(x)=ex−1+e1−x+x2−2x+af(x)=e^{x-1}+e^{1-x}+x^{2}-2x+af(x)=ex−1+e1−x+x2−2x+a,(1)证明:f(x)f(x)f(x)关于x=1x=1x=1对称;(2)若f(x)f(x)f(x)的最小值为3(ⅰ)求aaa;(ⅱ)不等式f(m(ex+e−x)+1)>f(ex−e−x)f(m(e^{x}+e^{-x})+1)>f(e^{x}-e^{-x})f(m(ex+e−x)+1)>f(ex−e−x)恒成立,求mmm的取值范围