Đáp án: $m = 0;m = - 1$
Giải thích các bước giải:
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = 2m + 1
\end{array} \right.\\
Khi:x_1^2 = {x_2}\\
\Leftrightarrow {x_1}.x_1^2 = 2m + 1\\
\Leftrightarrow x_1^3 = 2m + 1\\
\Leftrightarrow {x_1} = \sqrt[3]{{2m + 1}}\\
\Leftrightarrow {x_2} = {\left( {\sqrt[3]{{2m + 1}}} \right)^2}\\
\Leftrightarrow \sqrt[3]{{2m + 1}} + {\left( {\sqrt[3]{{2m + 1}}} \right)^2} = 2\left( {m + 1} \right)\\
\Leftrightarrow \sqrt[3]{{2m + 1}} + {\left( {\sqrt[3]{{2m + 1}}} \right)^2} = 2m + 1 + 1\\
Dat:\sqrt[3]{{2m + 1}} = a\\
\Leftrightarrow a + {a^2} = {a^3} + 1\\
\Leftrightarrow {a^3} - {a^2} - a + 1 = 0\\
\Leftrightarrow \left( {a - 1} \right)\left( {{a^2} - 1} \right) = 0\\
\Leftrightarrow {\left( {a - 1} \right)^2}\left( {a + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = 1\\
a = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt[3]{{2m + 1}} = 1\\
\sqrt[3]{{2m + 1}} = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2m + 1 = 1 \Leftrightarrow m = 0\\
2m + 1 = - 1 \Leftrightarrow m = - 1
\end{array} \right.\\
Vậy\,m = 0;m = - 1
\end{array}$
