a, \(\Delta=\left(-3\right)^2-4.1.\left(-5\right)=9+20=29>0\)
Suy ra pt luôn có 2 nghiệm phân biệt
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=-5\end{matrix}\right.\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=3^2-2.\left(-5\right)=9+10=19\)
\(b,A=\dfrac{x_1-2}{x_2}+\dfrac{x_2-2}{x_1}\\ =\dfrac{x_1\left(x_1-2\right)+x_2\left(x_2-2\right)}{x_1x_2}\\ =\dfrac{x_1^2-2x_1+x_2^2-2x_2}{-5}\\ =\dfrac{\left(x_1+x_2\right)^2-2\left(x_1+x_2\right)}{-5}\\ =\dfrac{3^2-2.3}{-5}\\ =\dfrac{-3}{5}\)