a: \(a=3;b=-5;c=-1\)
\(\text{Δ}=b^2-4ac=\left(-5\right)^2-4\cdot3\cdot\left(-1\right)=25+12=37\)
b:
b1:
Theo vi-et, ta có;
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{5}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)
b2:
\(A=\dfrac{2}{5x_1}+\dfrac{2}{5x_2}=\dfrac{2}{5}\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)\)
\(=\dfrac{2}{5}\cdot\dfrac{x_1+x_2}{x_1x_2}\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{5}{3}:\dfrac{-1}{3}\right)=\dfrac{2}{5}\cdot\left(-5\right)=-2\)
\(B=\dfrac{2024}{x_1}+\dfrac{2024}{x_2}=2024\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)=2024\cdot\left(-5\right)=-10120\)