26.(2023春•朝阳区期末)已知函数f(x)=e2xf(x)=e^{2x}f(x)=e2x,g(x)=m(2x+1)(m∈R)g(x)=m(2x+1)(m\in R)g(x)=m(2x+1)(m∈R).(Ⅰ)当m=1m=1m=1时,证明f(x)⩾g(x)f(x)\geqslant g(x)f(x)⩾g(x);(Ⅱ)若直线y=g(x)y=g(x)y=g(x)是曲线y=f(x)y=f(x)y=f(x)的切线,设h(x)=f(x)−g(x)h(x)=f(x)-g(x)h(x)=f(x)−g(x),求证:对任意的a>ba>ba>b,都有h(a)−h(b)a−b<2e2a−2\frac{h(a)-h(b)}{a-b}<2{e}^{2a}-2a−bh(a)−h(b)<2e2a−2.