Đáp án:
\(\dfrac{{24}}{{25}}\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\Delta ' = 1 - 5.\left( { - 1} \right) = 6\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{2}{5}\\
{x_1}{x_2} = - \dfrac{1}{5}
\end{array} \right.\\
{\left( {{x_1} - {x_2}} \right)^2} = {x_1}^2 - 2{x_1}{x_2} + {x_2}^2\\
= {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}\\
= {\left( { - \dfrac{2}{5}} \right)^2} - 4.\left( { - \dfrac{1}{5}} \right) = \dfrac{{24}}{{25}}
\end{array}\)
