Cho hàm số y=f(x)=e^1/x(x+1). Tính

* Hướng dẫn giải

Đáp án B

Ta có $f\left(x\right)=e^{\frac{1}{x\left(x+1\right)}}\iff f\left(x\right)=\frac{e^{\frac{1}{x}}}{e^{\frac{1}{x+1}}}$

$\Rightarrow T=f\left(1\right).f\left(2\right)\mathrm{...}f\left(2017\right).\sqrt[2018]{e}=\frac{e}{e^{\frac{1}{2}}}.\frac{e^{\frac{1}{2}}}{e^{\frac{1}{3}}}\mathrm{...}\frac{e^{\frac{1}{2017}}}{e^{\frac{1}{2018}}}.e^{\frac{1}{2018}}\iff T=e$