Đáp án:
m=0
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 1 - m + 1 \ge 0\\
\to 2 \ge m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
{x_1}^4 - {x_1}^3 = {x_2}^4 - {x_2}^3\\
\to {x_1}^4 - {x_2}^4 = {x_1}^3 - {x_2}^3\\
\to \left( {{x_1}^2 - {x_2}^2} \right)\left( {{x_1}^2 + {x_2}^2} \right) = \left( {{x_1} - {x_2}} \right)\left( {{x_1}^2 + {x_1}{x_2} + {x_2}^2} \right)\\
\to \left( {{x_1} - {x_2}} \right)\left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 + {x_2}^2} \right) = \left( {{x_1} - {x_2}} \right)\left( {{x_1}^2 + {x_1}{x_2} + {x_2}^2} \right)\\
\to \left( {{x_1} - {x_2}} \right)\left[ {2\left( {{x_1}^2 + {x_2}^2} \right) - {x_1}^2 - {x_1}{x_2} - {x_2}^2} \right] = 0\\
\to \left[ \begin{array}{l}
{x_1} - {x_2} = 0\\
{x_1}^2 + {x_2}^2 - {x_1}{x_2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x_1} = {x_2}\\
{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 3{x_1}{x_2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\Delta ' = 0\\
{\left( {{x_1} + {x_2}} \right)^2} - 3{x_1}{x_2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
2 - m = 0\\
4 - 3\left( {m - 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 0\left( {TM} \right)\\
m = \dfrac{7}{3}\left( {KTM} \right)
\end{array} \right.
\end{array}\)
