Đáp án:
b. \(m \in \left( {2; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\frac{{{x^2} + 2mx - 4}}{{2{x^2} - x + 2}} < 1\\
\to \frac{{{x^2} + 2mx - 4 - 2{x^2} + x - 2}}{{2{x^2} - x + 2}} < 0\\
\to \frac{{ - {x^2} + \left( {2m + 1} \right)x - 6}}{{2{x^2} - x + 2}} < 0\forall x\\
Do:2{x^2} - x + 2 > 0\forall x \in R\\
\to - {x^2} + \left( {2m + 1} \right)x - 6 < 0\\
\to 4{m^2} + 4m + 1 - 4.6 < 0\\
\to 4{m^2} + 4m - 23 < 0\\
\to x \in \left( {\frac{{ - 1 - 2\sqrt 6 }}{2};\frac{{ - 1 + 2\sqrt 6 }}{2}} \right)\\
b.(m - 1){x^2} + 2(m + 1)x + 4m + 1 > 0\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 1 > 0\\
{m^2} + 2m + 1 - \left( {m - 1} \right)\left( {4m + 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 1\\
{m^2} + 2m + 1 - 4{m^2} + 3m + 1 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 1\\
- 3{m^2} + 5m + 2 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 1\\
\left( {2 - m} \right)\left( {3m + 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 1\\
m \in \left( { - \infty ; - \frac{1}{3}} \right) \cup \left( {2; + \infty } \right)
\end{array} \right.\\
KL:m \in \left( {2; + \infty } \right)
\end{array}\)
