Hàm số đã cho xác định và liên tục trên $\left[1;2\right]$ .

$y\text{'}=\frac{1-m}{{\left(x+1\right)}^2}$

+ Nếu $m=1\Rightarrow y=1\Rightarrow \underset{\left[1;2\right]}{\max }y=\underset{\left[1;2\right]}{\min }y=1\Rightarrow \underset{\left[1;2\right]}{\max }y+\underset{\left[1;2\right]}{\min }y=2$  (loại).

+ Nếu $m\ne 1\Rightarrow$  Hàm số đơn điệu trên $\left[1;2\right]$

$\Rightarrow \left[\begin{array}{l}\left\{\begin{array}{l}\underset{\left[1;2\right]}{\min }y=y\left(1\right)\\ \underset{\left[1;2\right]}{\max }y=y\left(2\right)\end{array}\right.\\ \left\{\begin{array}{l}\underset{\left[1;2\right]}{\min }y=y\left(2\right)\\ \underset{\left[1;2\right]}{\max }y=y\left(1\right)\end{array}\right.\end{array}\right.\Rightarrow \underset{\left[1;2\right]}{\min }y+\underset{\left[1;2\right]}{\max }y=y\left(1\right)+y\left(2\right)=\frac{1+m}{2}+\frac{2+m}{3}=\frac{7+5m}{6}$.