已知函数\(f(x)=(ax+{\ln}x)\cdot(x-{\ln}x)-x^2\)有三个不同的零点\(x_1,x_2,x_3\) \((\)其中\(x_1<x_2<x_3)\),则\(\left(1-\dfrac{{\ln}x_1}{x_1}\right)^2\left(1-\dfrac{{\ln}x_2}{x_2}\right)\left(1-\dfrac{{\ln}x_3}{x_3}\right)\)的值为\((\qquad)\)
\(\mathrm{A}. 1-a\) \(\qquad \mathrm{B}.a-1\) \(\qquad \mathrm{C}.-1\) \(\qquad \mathrm{D}.1\)
若记$$t(x)=\dfrac{{\ln}x}{x},x>0.$$则\(t\)的取值范围为\(\left(-\infty,\dfrac{1}{\mathrm{e}}\right]\),则由\(f(x)=0\)可得关于\(t\)的一元二次方程$$
t^2=(1-a)(t-1),t\leqslant \dfrac{1}{\mathrm{e}}.$$如下图绘制出二次函数\(y=t^2,t\leqslant \dfrac{1}{\mathrm{e}}\)与\(y=(1-a)(t-1)\)的图象,若要使得\(f(x)\)有三个不同的零点,
t_1<0<t_2,t_1+t_2=1-a,t_1t_2=1-a.$$再结合\(t(x)=\dfrac{{\ln}x}{x}\)的图象,可知$$t_1=t(x_1),t_2=t(x_2)=t(x_3).$$
(1-t_1)2(1-t_2)2=\left[1-(t_1+t_2)+t_1t_2\right]^2=1.$$
