$y= \dfrac{x+m}{x+1}$. TXĐ: $D= \mathbb{R} \setminus \{ -1 \}$
$y'=\dfrac{x+1-(x+m)}{(x+1)^2} = \dfrac{1-m}{(x+1)^2}$
+ Khi $m=m_0$: $\displaystyle \min_{[1;2]}y=y(1)= \dfrac{1+m_0}{2}$,
$\displaystyle \max_{[1;2]}y=y(2)= \dfrac{2+m_0}{3}$
Theo đề ta có: $\displaystyle \min_{[1;2]}y + \max_{[1;2]}y= \dfrac{16}{3} \\
\Leftrightarrow \dfrac{1+m_0}{2} + \dfrac{2+m_0}{3} = \dfrac{16}{3} \\
\Leftrightarrow 3(1+m_0)+2(2+m_0)=32 \\
\Leftrightarrow m_0=5$