Chọn phương án A.

Ta có \(y'=\dfrac{2}{(x+1)^2}\).

Theo đề bài thì $$\begin{aligned}y'\left(x_0\right)=\dfrac{1}{2}\Leftrightarrow&\dfrac{2}{\left(x_0+1\right)^2}=\dfrac{1}{2}\\ \Leftrightarrow&\left[\begin{array}{l}x_0+1=2\\ x_0+1=-2\end{array}\right.\\ \Leftrightarrow&\left[\begin{array}{l}x_0=1\\ x_0=-3.\end{array}\right.\end{aligned}$$

• \(x_0=1\Rightarrow y_0=0\)
  \(\Rightarrow\) Tiếp tuyến: \(y=\dfrac{1}{2}(x-1)+0=\dfrac{1}{2}x-\dfrac{1}{2}\).
• \(x_0=-3\Rightarrow y_0=2\)
  \(\Rightarrow\) Tiếp tuyến: \(y=\dfrac{1}{2}(x+3)+2=\dfrac{1}{2}x+\dfrac{7}{2}\).