Đáp án:
d) \(m > - \dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} - 4x + m - 5 > 0\forall x\\
\to 4 - m + 5 < 0\\
\to 9 < m\\
b){x^2} - \left( {m + 2} \right)x + 8m + 1 > 0\forall x\\
\to {m^2} + 4m + 4 - 4\left( {8m + 1} \right) < 0\\
\to {m^2} + 4m + 4 - 32m - 4 < 0\\
\to {m^2} - 28m < 0\\
\to m\left( {m - 28} \right) < 0\\
\to 0 < m < 28\\
c){x^2} + 4x + {\left( {m - 2} \right)^2} > 0\forall x\\
\to 4 - {\left( {m - 2} \right)^2} < 0\\
\to 4 - {m^2} + 4m - 4 < 0\\
\to - m\left( {m - 4} \right) < 0\\
\to \left[ \begin{array}{l}
m > 4\\
m < 0
\end{array} \right.\\
d)\left( {3m + 1} \right){x^2} - \left( {3m + 1} \right)x + m + 4 > 0\forall x\\
\to \left\{ \begin{array}{l}
3m + 1 > 0\\
9{m^2} + 6m + 1 - 4\left( {3m + 1} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{1}{3}\\
9{m^2} + 6m + 1 - 4\left( {3{m^2} + 13m + 4} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{1}{3}\\
9{m^2} + 6m + 1 - 12{m^2} - 52m - 16 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{1}{3}\\
- 3{m^2} - 46m - 15 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{1}{3}\\
\left[ \begin{array}{l}
m > - \dfrac{1}{3}\\
m < - 15
\end{array} \right.
\end{array} \right.\\
\to m > - \dfrac{1}{3}
\end{array}\)
