Giúp em bài 12 câu b,c với ạ

Đáp án:

$\begin{array}{l}
b){x^2} + 2\left( {m + 1} \right).x + m + 7 > 0\forall x\\
 \Rightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta ' < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
1 > 0\left( {tmdk} \right)\\
{\left( {m + 1} \right)^2} - m - 7 < 0
\end{array} \right.\\
 \Rightarrow {m^2} + 2m + 1 - m - 7 < 0\\
 \Rightarrow {m^2} + m - 6 < 0\\
 \Rightarrow \left( {m - 2} \right)\left( {m + 3} \right) < 0\\
 \Rightarrow  - 3 < m < 2\\
Vậy\, - 3 < m < 2\\
c)\left( {m - 1} \right){x^2} - 2\left( {m + 1} \right).x + 3\left( {m - 2} \right) < 0\forall x\\
 \Rightarrow \left\{ \begin{array}{l}
m - 1 < 0\\
\Delta ' < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 1\\
{\left( {m + 1} \right)^2} - 3.\left( {m - 1} \right)\left( {m - 2} \right) < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 1\\
{m^2} + 2m + 1 - 3{m^2} + 9m - 6 < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 1\\
2{m^2} - 11m + 5 > 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 1\\
\left( {2m - 1} \right)\left( {m - 5} \right) > 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 1\\
\frac{1}{2} < m < 5
\end{array} \right.\\
 \Rightarrow \frac{1}{2} < m < 1\\
Vậy\,\frac{1}{2} < m < 1
\end{array}$