Cho hàm số

- Hướng dẫn giải

$$\begin{gathered} 1\left)f\right(x)=\frac{1}{3+2^x}+\frac{1}{3+2^x} \\ =\frac{1}{3+2^x}+\frac{2^x}{3.2^x+1}=\frac{4^x+6.2^x+1}{3.4^x+10.2^x+3} \end{gathered}$$

$\Rightarrow f\text{'}\left(x\right)=\frac{\left(2.4^x.\ln 2+5.2^x.\ln 2\right)\left(3.4^x+10.2^x+3\right)}{{\left(3.4^x+10.2^x+3\right)}^2}$

$-\frac{\left(6.4^x.\ln 2+10.2^x.\ln 2\right)\left(4^x+6.2^x+1\right)}{{\left(3.4^x+10.2^x+3\right)}^2}$

$$\begin{gathered} =\frac{\left(2.2^x+6\right)\left(3.4^x+10.2^x+3\right)-\left(6.2^x+10\right)\left(4^x+6.2^x+1\right)}{{\left(3.4^x+10.2^x+3\right)}^2} \\ .2^x.\ln 2=\frac{-8.4^x+8}{{\left(3.4^x+10.2^x+3\right)}^2}.2^x.\ln 2 \end{gathered}$$

$f\text{'}\left(x\right)=0\iff -8.4^x+8=0\iff 4^x=1\iff x=0$

$2\left)f\right(x)=\frac{4^x+6.2^x+1}{3.4^x+10.2^x+3}$

Ta có

$$\begin{gathered} f\left(x\right)-\frac{1}{3}=\frac{4^x+6.2^x+1}{3.4^x+10.2^x+3}-1 \\ =\frac{-2.4^x-4.2^x-2}{3.4^x+10.2^x+3}<0,\forall x \\ \Rightarrow f\left(1\right)+f\left(2\right)+..+f\left(2017\right) \\ <1+1+...+1=2017 \\ \Rightarrow f\left(1\right)+f\left(2\right)+..+f(2017=2017 \\ \Rightarrow 2) sai \end{gathered}$$

3)$$\begin{gathered} f\left(x^2\right)=\frac{1}{3+2^x}+\frac{1}{3+2^{-x}} \\ \Rightarrow f\left(x^2\right)=\frac{1}{3+4^x}+\frac{1}{3+4^{-x}} là sai \end{gathered}$$

Chọn đáp án A.