Đáp án:
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 1\\
{x_1}.{x_2} = - 2
\end{array} \right.\\
A = \dfrac{{{x_1} - 2{x_2}}}{{{x_2} + 2}} + \dfrac{{{x_2} - 2{x_1}}}{{{x_1} + 2}} + 5\\
= \dfrac{{{x_1} + 4 - 2\left( {{x_2} + 2} \right)}}{{{x_2} + 2}} + \dfrac{{{x_2} + 4 - 2\left( {{x_1} + 2} \right)}}{{{x_1} + 2}} + 5\\
= \dfrac{{{x_1} + 4}}{{{x_2} + 2}} - 2 + \dfrac{{{x_2} + 4}}{{{x_1} + 2}} - 2 + 5\\
= \dfrac{{\left( {{x_1} + 4} \right)\left( {{x_1} + 2} \right) + \left( {{x_2} + 4} \right)\left( {{x_2} + 2} \right)}}{{\left( {{x_1} + 2} \right)\left( {{x_2} + 2} \right)}} + 1\\
= \dfrac{{x_1^2 + x_2^2 + 6\left( {{x_1} + {x_2}} \right) + 16}}{{{x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right) + 4}} + 1\\
= \dfrac{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2} + 6.1 + 16}}{{ - 2 + 2 + 4}} + 1\\
= \dfrac{{1 - 2.\left( { - 2} \right) + 6 + 16}}{4} + 1\\
= \dfrac{{31}}{4}
\end{array}$
