cho hàm số fx)= a^x/(a^x+căn a) với a>0, a khác 1

* Hướng dẫn giải

Đáp án C

Ta có

 $\begin{array}{l}f\left(x\right)=\frac{a^x}{a^x+\sqrt{a}}\\ \Rightarrow f\left(1-x\right)=\frac{a^{1-x}}{a^{1-x}+\sqrt{a}}\\ =\frac{a}{a^x\left(\frac{a}{a^x}+\sqrt{a}\right)}\\ =\frac{a}{\sqrt{a}\left(a^x+\sqrt{a}\right)}\\ =\frac{\sqrt{a}}{a^x+\sqrt{a}}\end{array}$

 $\begin{array}{l}\Rightarrow f\left(x\right)+f\left(1-x\right)\\ =\frac{a^x}{a^x+\sqrt{a}}+\frac{\sqrt{a}}{a^x+\sqrt{a}}=1\end{array}$

Áp dụng vào bài toán ta có:

$P=f\left(\frac{1}{2018}\right)+f\left(\frac{2}{2018}\right)+\mathrm{...}+f\left(\frac{2017}{2018}\right)$

$=\left[f\left(\frac{1}{2018}\right)+f\left(\frac{2017}{2018}\right)\right]$$+\left[f\left(\frac{2}{2018}\right)+f\left(\frac{2016}{2018}\right)\right]+\mathrm{...}+$$\left[f\left(\frac{1007}{2018}\right)+f\left(\frac{1011}{2018}\right)\right]$$+\left[f\left(\frac{1008}{2018}\right)+f\left(\frac{1010}{2018}\right)\right]$$+f\left(\frac{1009}{2018}\right)$

$\begin{array}{l}=1.1008+f\left(\frac{1009}{2018}\right)\\ =1008+\frac{a^{\frac{1}{2}}}{a^{\frac{1}{2}}+\sqrt{a}}\\ =\mathrm{1008,5}=\frac{2017}{2}\end{array}$