Đáp án:

 $a)m>1\\ b)m\in \varnothing\\ c)0<m<1\\ d)m=\dfrac{-5\pm\sqrt{41}}{8}$

Giải thích các bước giải:

$ f(x)=\dfrac{1}{3}(m-1)x^3-mx^2+(m+2)x-5\\ f'(x)=(m-1)x^2-2mx+m+2\\ a)f'(x) \ge 0 \, \forall \, x\\ \Leftrightarrow \left\{\begin{array}{l} \Delta' \le 0\\m-1 >0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m^2-(m-1)(m+2)\le 0\\m-1 >0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2-m\le 0\\m-1 >0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ge 2\\m>1\end{array} \right.\\ \Leftrightarrow m>1\\ b)f'(x) \le 0 \, \forall \, x\\ \Leftrightarrow \left\{\begin{array}{l} \Delta' \le 0\\m-1 <0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2-m\le 0\\m-1<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ge 2\\m<1\end{array} \right.\\ \Leftrightarrow m\in \varnothing$

$c)f'(x)$ có hai nghiệm cùng âm

$\Leftrightarrow \left\{\begin{array}{l} \Delta'> 0\\\dfrac{2m}{m-1} <0\\\dfrac{m+2}{m-1}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2-m> 0\\\dfrac{2m}{m-1} <0\\\dfrac{m+2}{m-1}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m<2\\0<m<1\\-2<m<1\end{array} \right.\\ \Leftrightarrow 0<m<1\\ d)\Delta' >0 \Leftrightarrow m<2\\ \circledast x_1>x_2\\ x_1=\dfrac{m+\sqrt{2-m}}{m-1}(m/ne 1)\\ x_2=\dfrac{m-\sqrt{2-m}}{m-1}(m/ne 1)\\ x_1+2x_2=1\\ \Leftrightarrow \dfrac{m+\sqrt{2-m}}{m-1}+2\dfrac{m-\sqrt{2-m}}{m-1}=1\\ \Leftrightarrow  \dfrac{3m-\sqrt{2-m}}{m-1}=1\\ \Leftrightarrow 3m-\sqrt{2-m}=m-1\\ \Leftrightarrow 2m+1=\sqrt{2-m}\left(m \ge -\dfrac{1}{2}\right)\\ \Leftrightarrow (2m+1)^2=2-m\\ \Leftrightarrow m=\dfrac{-5+\sqrt{41}}{8}(TM);m=\dfrac{-5-\sqrt{41}}{8}(L)\\ \circledast x_1<x_2\\ x_1+2x_2=1\\ \Leftrightarrow \dfrac{m-\sqrt{2-m}}{m-1}+2\dfrac{m+\sqrt{2-m}}{m-1}=1\\ \Leftrightarrow  \dfrac{3m+\sqrt{2-m}}{m-1}=1\\ \Leftrightarrow 3m+\sqrt{2-m}=m-1\\ \Leftrightarrow -2m-1=\sqrt{2-m}\left(m \le -\dfrac{1}{2}\right)\\ \Leftrightarrow (2m+1)^2=2-m\\ \Leftrightarrow m=\dfrac{-5+\sqrt{41}}{8}(L);m=\dfrac{-5-\sqrt{41}}{8}(TM)$

Kết hợp hai trường hợp, $m=\dfrac{-5\pm\sqrt{41}}{8}$