Giải thích các bước giải:
$\begin{array}{l}
a)\left( {m - 1} \right){x^2} + 2mx + 3m - 2 \ge 0,\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 1 > 0\\
\Delta ' \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
{m^2} - \left( {m - 1} \right)\left( {3m - 2} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
- 2{m^2} + 5m - 2 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
2{m^2} - 5m + 2 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
\left( {2m - 1} \right)\left( {m - 2} \right) \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
\left[ \begin{array}{l}
m \ge 2\\
m \le \dfrac{1}{2}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow m \ge 2
\end{array}$
Vậy $m \in \left[ {2; + \infty } \right)$ thỏa mãn đề.
$\begin{array}{l}
b)m{x^2} - 2mx + 3 - 2m > 0,\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\Delta ' < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{\left( { - m} \right)^2} - m.\left( {3 - 2m} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
3{m^2} - 3m < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
3m\left( {m - 1} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
0 < m < 1
\end{array} \right.\\
\Leftrightarrow 0 < m < 1
\end{array}$
Vậy $m \in \left( {0;1} \right)$ thỏa mãn đề
