Cho hàm số f(x) có đạo hàm trên R , biết f'(x)-2018f9x)=2018.2017.x^2017.e^2018x với mọi x thuộc R; f(0)=2018 . Giá trị của f(1) là

Đáp án D

Ta có: $f^{\text{'}}\left(x\right)-2018f\left(x\right)=2018.x^{2017}.e^{2018\text{x}}\iff f^{\text{'}}\left(x\right).e^{-2018\text{x}}-2018{\text{e}}^{-2018\text{x}}f\left(x\right)=2018{\text{x}}^{2017}$

$\Rightarrow {\left[f\left(x\right)e^{-2018\text{x}}\right]}^{\text{'}}={\left(x^{2018}\right)}^{\text{'}}\Rightarrow \int {[f\left(x\right).e^{-2018\text{x}}]}^{\text{'}}d\text{x}=\int {\left(x^{2018}\right)}^{\text{'}}d\text{x}\Rightarrow f\left(x\right)e^{-2018\text{x}}=x^{2018}+C$Do $f\left(0\right)=2018\Rightarrow f\left(0\right).e^0=C\Rightarrow C=2018\Rightarrow f\left(x\right)=x^{2018}.e^{2018\text{x}}+2018{\text{e}}^{2018\text{x}}$

$\Rightarrow f\left(1\right)=e^{2018}+2018{\text{e}}^{2018}=2019{\text{e}}^{2018\text{x}}$.