Đáp án:
d) \(m < \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
b)\left\{ \begin{array}{l}
{m^2} - 1 > 0\\
{m^2} + 2m + 1 - 3\left( {{m^2} - 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
{m^2} + 2m + 1 - 3{m^2} + 3 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
- 2{m^2} + 2m + 4 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
\left[ \begin{array}{l}
m > 2\\
m < - 1
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
m > 2\\
m < - 1
\end{array} \right.\\
c)\left\{ \begin{array}{l}
{m^2} + 3 > 0\left( {ld} \right)\forall m\\
{m^2} + 2m + 1 - 1.\left( {{m^2} + 3} \right) < 0
\end{array} \right.\\
\to {m^2} + 2m + 1 - {m^2} - 3 < 0\\
\to 2m - 2 < 0\\
\to m < 1\\
d)\left\{ \begin{array}{l}
{m^2} + 2 > 0\left( {ld} \right)\forall m\\
{m^2} + 2m + 1 - 1\left( {{m^2} + 2} \right) < 0
\end{array} \right.\\
\to {m^2} + 2m + 1 - {m^2} - 2 < 0\\
\to 2m - 1 < 0\\
\to m < \dfrac{1}{2}
\end{array}\)
