30.(2023春•天津期末)已知函数f(x)=(a−1)lnx−(a−1)x+1f(x)=(a-1)lnx-(a-1)x+1f(x)=(a−1)lnx−(a−1)x+1.(1)证明:当a=2a=2a=2时,f(x)⩽0f(x)\leqslant 0f(x)⩽0恒成立;(2)若g(x)=f(x)+x22−x−1+a(a>2)g(x)=f(x)+\frac{{x}^{2}}{2}-x-1+a(a>2)g(x)=f(x)+2x2−x−1+a(a>2)且g(m)=12(m≠1)g(m)=\frac{1}{2}(m\ne 1)g(m)=21(m=1),证明:∀x∈(1\forall x\in (1∀x∈(1,m]m]m],m<2a−3m<2a-3m<2a−3.