Đáp án:

$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} =  - 4
\end{array} \right.\\
a)A = x_1^2 + x_2^2 - {x_1}{x_2}\\
 = x_1^2 + 2{x_1}{x_2} + x_2^2 - 3{x_1}{x_2}\\
 = {\left( {{x_1} + {x_2}} \right)^2} - 3{x_1}{x_2}\\
 = {2^2} - 3.\left( { - 4} \right)\\
 = 4 + 12\\
 = 16\\
b)B = x_1^3 + x_2^3 - {x_1} - {x_2}\\
 = \left( {{x_1} + {x_2}} \right)\left( {x_1^2 - {x_1}{x_2} + x_2^2} \right) - \left( {{x_1} + {x_2}} \right)\\
 = 2.16 - 2\\
 = 30\\
c)C = \left( {{x_1} + 2} \right)\left( {{x_2} + 2} \right)\\
 = {x_1}{x_2} + 2{x_1} + 2{x_2} + 4\\
 =  - 4 + 2.\left( {{x_1} + {x_2}} \right) + 4\\
 =  - 4 + 2.2 + 4\\
 = 4\\
d)D = {\left( {{x_1} + 1} \right)^2} + {\left( {{x_2} + 1} \right)^2}\\
 = x_1^2 + 2{x_1} + 1 + x_2^2 + 2{x_2} + 1\\
 = {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + 2.\left( {{x_1} + {x_2}} \right) + 2\\
 = {2^2} - 2.\left( { - 4} \right) + 2.2 + 2\\
 = 4 + 8 + 4 + 2\\
 = 16\\
e)E = \frac{{x_1^3}}{{{x_2}}} + \frac{{x_2^3}}{{{x_1}}}\\
 = \frac{{x_1^4 + x_2^4}}{{{x_1}{x_2}}}\\
 = \frac{{{{\left( {x_1^2 + x_2^2} \right)}^2} - 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{ - 4}}\\
 = \frac{{{{\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right]}^2} - 2.{{\left( { - 4} \right)}^2}}}{{ - 4}}\\
 = \frac{{{{\left( {{2^2} - 2.\left( { - 4} \right)} \right)}^2} - 2.16}}{{ - 4}}\\
 = \frac{{144 - 32}}{{ - 4}}\\
 =  - 28\\
F = \frac{1}{{{x_1} - 2}} + \frac{1}{{{x_2} - 2}}\\
 = \frac{{{x_2} - 2 + {x_1} - 2}}{{\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right)}}\\
 = \frac{{2 - 4}}{{{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4}}\\
 = \frac{{ - 2}}{{ - 4 - 2.2 + 4}}\\
 = \frac{{ - 2}}{{ - 4}} = \frac{1}{2}\\
G = x_1^{10} + x_2^{10} - 2\left( {x_1^9 + x_2^9} \right) - 4\left( {x_1^8 + x_2^8} \right)\\
 = x_1^8\left( {x_1^2 - 2{x_1} - 4} \right) + x_2^8\left( {x_2^2 - 2{x_2} - 4} \right)\\
 = x_1^8.0 + x_2^8.0\\
 = 0
\end{array}$