Ta có pt: \(x^2-x-1=0\left(a=1;b=-1;c=-1\right)\)
Theo vi-et ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}==\dfrac{-\left(-1\right)}{1}=1\\x_1x_2=\dfrac{c}{a}=\dfrac{-1}{1}=-1\end{matrix}\right.\)
Biểu thức: \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-1}=-1\)

