Đáp án:
Phương trình bậc 2 có dạng $a.{x^2} + bx + c = 0$ có 2 nghiệm khi và chỉ khi: $\left\{ \begin{array}{l}
a \ne 0\\
\Delta \ge 0
\end{array} \right.$
$\begin{array}{l}
5)\left\{ \begin{array}{l}
m \ne 0\\
{\left( {m + 1} \right)^2} - m\left( {m - 2} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^2} + 2m + 1 - {m^2} + 2m \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
m \ge - \dfrac{1}{4}
\end{array} \right.\\
\text{Vậy}\,m \ge - \dfrac{1}{4};m \ne 0\\
6)\left\{ \begin{array}{l}
m \ne 0\\
{\left( {3 - m} \right)^2} - 4m\left( {\dfrac{m}{4} - 2} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 6m + 9 - {m^2} + 8m \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
2m + 9 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
m \ge - \dfrac{9}{2}
\end{array} \right.\\
\text{Vậy}\,m \ge - \dfrac{9}{2};m \ne 0\\
7)\left\{ \begin{array}{l}
m \ne 1\\
{\left( {2m - 1} \right)^2} - 4\left( {m - 1} \right)\left( {m + 2} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
4{m^2} - 4m + 1 - 4{m^2} - 4m + 8 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
m \le \dfrac{9}{8}
\end{array} \right.\\
\text{Vậy}\,m \le \dfrac{9}{8};m \ne 1\\
8)\left\{ \begin{array}{l}
m \ne - \dfrac{1}{2}\\
{\left( {2m - 3} \right)^2} - \left( {2m - 1} \right)\left( {2m + 1} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - \dfrac{1}{2}\\
4{m^2} - 12m + 9 - \left( {4{m^2} - 1} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - \dfrac{1}{2}\\
m \le \dfrac{5}{6}
\end{array} \right.\\
\text{Vậy}\,m \le \dfrac{5}{6};m \ne - \dfrac{1}{2}\\
9)\left\{ \begin{array}{l}
m \ne 1\\
m \ge 1
\end{array} \right.\\
\Rightarrow m > 1\\
10)\left\{ \begin{array}{l}
m \ne \dfrac{3}{2}\\
m \le \dfrac{3}{2}
\end{array} \right.\\
\Rightarrow m < \dfrac{3}{2}
\end{array}$
