Đáp án:
$\begin{array}{l}
1)f\left( x \right) > 0\forall x\\
\Leftrightarrow m.{x^2} - 2\left( {m - 1} \right).x + 4m - 1 > 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\Delta ' < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{\left( {m - 1} \right)^2} - m.\left( {4m - 1} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{m^2} - 2m + 1 - 4{m^2} + m < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
3{m^2} + m - 1 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m > \dfrac{{ - 1 + \sqrt {13} }}{6}\\
m < \dfrac{{ - 1 - \sqrt {13} }}{6}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow m > \dfrac{{\sqrt {13} - 1}}{6}\\
Vậy\,m > \dfrac{{\sqrt {13} - 1}}{6}\\
2)a)5{x^2} - x + m > 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
5 > 0\left( {tm} \right)\\
\Delta < 0
\end{array} \right.\\
\Leftrightarrow 1 - 4.5.m < 0\\
\Leftrightarrow m > \dfrac{1}{{20}}\\
Vậy\,m > \dfrac{1}{{20}}\\
b)m\left( {m + 2} \right).{x^2} + 2mx + 2 > 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
m\left( {m + 2} \right) > 0\\
\Delta ' < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 0\\
m < - 2
\end{array} \right.\\
{m^2} - 2m\left( {m + 2} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 0\\
m < - 2
\end{array} \right.\\
{m^2} + 4m > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 0\\
m < - 2
\end{array} \right.\\
\left[ \begin{array}{l}
m > 0\\
m < - 4
\end{array} \right.
\end{array} \right.\\
Vậy\,m < - 4\,hoac\,m > 0
\end{array}$
