Đáp án:
a) Phương trình có 2 nghiệm trái dấu
b) \(\dfrac{{172}}{9}\)
Giải thích các bước giải:
\(a)Xét:a.c = 3.\left( { - 1} \right) < 0\)
⇒ Phương trình có 2 nghiệm trái dấu
\(\begin{array}{l}
b)Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{2}{3}\\
{x_1}{x_2} = - \dfrac{1}{3}
\end{array} \right.\\
A = \left( {{x_1}^2 - 5{x_1} - 2} \right)\left( {{x_2}^2 + 5{x_2} - 2} \right) + 3\left( {{x_1}^2 + {x_2}^2} \right)\\
= {x_1}^2{x_2}^2 + 5{x_1}^2{x_2} - 2{x_1}^2 - 5{x_1}{x_2}^2 - 25{x_1}{x_2} + 10{x_1} - 2{x_2}^2 - 10{x_2} + 4 + 3{x_1}^2 + 3{x_2}^2\\
= {x_1}^2{x_2}^2 + 5{x_1}{x_2}\left( {{x_1} + {x_2}} \right) + {x_1}^2 + {x_2}^2 - 25{x_1}{x_2} + 10\left( {{x_1} + {x_2}} \right) + 4\\
= {x_1}^2{x_2}^2 + 5{x_1}{x_2}\left( {{x_1} + {x_2}} \right) + {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 27{x_1}{x_2} + 10\left( {{x_1} + {x_2}} \right) + 4\\
= {x_1}^2{x_2}^2 + 5{x_1}{x_2}\left( {{x_1} + {x_2}} \right) + {\left( {{x_1} + {x_2}} \right)^2} - 27{x_1}{x_2} + 10\left( {{x_1} + {x_2}} \right) + 4\\
= {\left( { - \dfrac{1}{3}} \right)^2} + 5.\left( { - \dfrac{1}{3}} \right).\dfrac{2}{3} + {\left( {\dfrac{2}{3}} \right)^2} - 27.\left( { - \dfrac{1}{3}} \right) + 10.\dfrac{2}{3} + 4\\
= \dfrac{{172}}{9}
\end{array}\)
