越复杂越简单，构造问题

已知函数\(f(x)=(\ln x-2x+a)\ln x\)
\((1)\)当\(a=2\)时，求\(f(x)\)的单调性
\((2)\)若\(f(x)\leq\dfrac{e^x}{x}-x^2+ax-a\)，求实数\(a\)的取值范围.
解
\((1)\) \(a=2,f(x)=(\ln x-2x+2)\ln x=\ln^2x-2x\ln x+2\ln x\)
\(f^{\prime}(x)=\dfrac{2\ln x}{x}-2\ln x-2+\dfrac{2}{x}=\dfrac{2\ln x-2x\ln x-2x+2}{x}=\dfrac{2(1-x)(\ln x+1)}{x}\)
则不难得到\(f(x)\)在\(\left(0,\dfrac{1}{e}\right)\)和\((1,+\infty)\)上单调递减，\(\left(\dfrac{1}{e},1\right)\)上单调递增

(2)\((\ln x-2x+a)\ln x\leq\dfrac{e^x}{x}-x^2+ax-a\)
即\(\ln^2x-2x\ln x+a\ln x\leq\dfrac{e^x}{x}-x^2+ax-a\)
即\(\ln^2x-2x\ln x+x^2\leq \dfrac{e^x}{x}+ax-a-a\ln x\)
即\((x-\ln x)^2\leq e^{x-\ln x}+a(x-\ln x-1)\)
记\(x-\ln x=t\)，则有\(t^2\leq e^t+a(t-1)(t\geq1)\)
即\(t^2-e^t\leq a(t-1)\)
即\(\dfrac{t^2-e^t}{t-1}\leq a\)记\(g(t)=\dfrac{t^2-e^t}{t-1}\)
\(g^{\prime}(t)=\dfrac{(t-1)(2t-e^t)-t^2+e^t}{(t-1)^2}=\dfrac{t^2+e^t(-t+2)-2t}{(t-1)^2}\)
在\(t\in(1,2)\)时，\(t^2+e^t(2-t)-2t>t^2-2t+(t+1)(2-t)=2-t>0\)
在\(t\in[2,+\infty)\)上\(t^2+e^t(2-t)-2t<t^2+(t+1)(2-t)-2t=2-t<0\)
从而\(a\geq g(2)=4-e^2\)