Đáp án: $m = - \dfrac{1}{2}$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} = x - m + 1\\
\Rightarrow {x^2} - x + m - 1 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow 1 - 4\left( {m - 1} \right) > 0\\
\Rightarrow 1 - 4m + 4 > 0\\
\Rightarrow m < \dfrac{5}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 1\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
A;B \in \left( P \right)\\
\Rightarrow \left\{ \begin{array}{l}
{y_1} = x_1^2\\
{y_2} = x_2^2
\end{array} \right.\\
{y_1} + {y_2} = 4\left( {{x_1} + {x_2}} \right)\\
\Rightarrow x_1^2 + x_2^2 = 4.1 = 4\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 4\\
\Rightarrow 1 - 2.\left( {m - 1} \right) = 4\\
\Rightarrow m - 1 = - \dfrac{3}{2}\\
\Rightarrow m = - \dfrac{1}{2}\left( {tmdk} \right)
\end{array}$
