Đáp án:
$\begin{array}{l}
a){x^2} - \left( {2m + 1} \right).x + {m^2} + 1 = 0\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {\left( {2m + 1} \right)^2} - 4\left( {{m^2} + 1} \right) \ge 0\\
\Rightarrow 4{m^2} + 4m + 1 - 4{m^2} - 4 \ge 0\\
\Rightarrow 4m \ge 3\\
\Rightarrow m \ge \dfrac{3}{4}\\
Vậy\,m \ge \dfrac{3}{4}\\
b)Dk:m > \dfrac{3}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 1\\
{x_1}{x_2} = {m^2} + 1
\end{array} \right.\\
Do:{x_1} = 2{x_2}\\
\Rightarrow \left\{ \begin{array}{l}
3{x_2} = 2m + 1\\
2x_2^2 = {m^2} + 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_2} = \dfrac{{2m + 1}}{3}\\
2x_2^2 = {m^2} + 1
\end{array} \right.\\
\Rightarrow 2.\dfrac{{2m + 1}}{3} = {m^2} + 1\\
\Rightarrow 3{m^2} - 4m + 1 = 0\\
\Rightarrow \left( {3m - 1} \right)\left( {m - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = \dfrac{1}{3}\left( {ktm} \right)\\
m = 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = 1
\end{array}$
