2020届成都二诊理科12题

该题可以划归为:\(\frac{\ln x_1}{x_1}=x_2 \text{e}^{-x_2}=k(k<0)\)

\(\Rightarrow \frac{\ln\frac{1}{ x_1}}{x_1}=-x_2 \text{e}^{-x_2}=-k(k<0)\)

\(\Rightarrow \ln(\ln\frac{1}{ x_1})+\ln(\frac{1}{x_1})=\ln(-x_2)+ (-x_2)=\ln(-k)(k<0)\)

构造函数\(f(x)=\ln(x)+x\)

\(\Rightarrow \ln\frac{1}{ x_1}=-x_2 \Rightarrow \ln x_1=x_2\)

\(\Rightarrow (\frac{x_2}{ x_1})^2\text{e}^k=(\frac{\ln x_1}{ x_1})^2\text{e}^{\frac{\ln x_1}{ x_1}}\)

构造函数\(y=x^2\text{e}^x\),其余略。

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转载自www.cnblogs.com/xuebajunlutiji/p/12660825.html