Đáp án: m=2
Giải thích các bước giải:
$\begin{array}{l}
c)\Delta ' \ge 0\\
\Rightarrow {\left( {m + 1} \right)^2} - {m^2} - 4 \ge 0\\
\Rightarrow {m^2} + 2m + 1 - {m^2} - 4 \ge 0\\
\Rightarrow m \ge \dfrac{3}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = {m^2} + 4
\end{array} \right.\\
x_2^2 - 2\left( {m + 1} \right){x_2} + {m^2} + 4 = 0\\
\Rightarrow x_2^2 - 2m{x_2} + {m^2} + 3 = 2{x_2} - 1\\
\Rightarrow \left( {2{x_1} - 1} \right).\left( {2{x_2} - 1} \right) = 21\\
\Rightarrow 4{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 1 = 21\\
\Rightarrow 4.\left( {{m^2} + 4} \right) - 2.2\left( {m + 1} \right) = 20\\
\Rightarrow {m^2} + 4 - m - 1 = 5\\
\Rightarrow {m^2} - m - 2 = 0\\
\Rightarrow \left( {m - 2} \right)\left( {m + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 2\left( {tmdk} \right)\\
m = - 1\left( {ktm} \right)
\end{array} \right.
\end{array}$
