Đáp án:
$\begin{array}{l}
{x^2} - 3x = 1\\
\Leftrightarrow {x^2} - 3x - 1 = 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = - 1
\end{array} \right.\\
A = {\left( {{x_1} - {x_2}} \right)^2}\\
= x_1^2 + 2{x_1}{x_2} + x_2^2 - 4{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}\\
= {3^2} - 4.\left( { - 1} \right)\\
= 9 + 4\\
= 13\\
B = \dfrac{{{x_1}}}{{{x_2}}} + \dfrac{{{x_2}}}{{{x_1}}}\\
= \dfrac{{x_1^2 + x_2^2}}{{{x_1}{x_2}}}\\
= \dfrac{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}}{{{x_1}{x_2}}}\\
= \dfrac{{{3^2} - 2.\left( { - 1} \right)}}{{ - 1}}\\
= \dfrac{{11}}{{ - 1}}\\
= - 11
\end{array}$