Đáp án:
$\begin{array}{l}
a)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{ - 5}}{2}\\
{x_1}{x_2} = - \frac{1}{2}
\end{array} \right.\\
b)x_1^2 + x_2^2\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {\left( { - \frac{5}{2}} \right)^2} - 2.\left( { - \frac{1}{2}} \right)\\
= \frac{{25}}{4} + 1 = \frac{{29}}{4}\\
c)x_1^3 + x_2^3\\
= {\left( {{x_1} + {x_2}} \right)^3} - 3{x_1}{x_2}\left( {{x_1} + {x_2}} \right)\\
= {\left( { - \frac{5}{2}} \right)^3} - 3.\left( { - \frac{1}{2}} \right).\left( { - \frac{5}{2}} \right)\\
= - \frac{{155}}{8}\\
d)x_1^2 - x_2^2\\
= \left( {{x_1} + {x_2}} \right)\left( {{x_1} - {x_2}} \right)\\
= \frac{{ - 5}}{2}.\sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\
= - \frac{{5\sqrt {33} }}{4}\\
e)x_1^3 - x_2^3\\
= \left( {{x_1} - {x_2}} \right)\left( {x_1^2 + {x_1}{x_2} + x_2^2} \right)\\
= \frac{{\sqrt {33} }}{2}.\left( {\frac{{29}}{4} - \frac{1}{2}} \right)\\
= \frac{{27\sqrt {33} }}{8}\\
g)x_1^4 - x_2^4\\
= \left( {x_1^2 + x_2^2} \right)\left( {x_1^2 - x_2^2} \right)\\
= \frac{{29}}{4}.\left( { - \frac{{5\sqrt {33} }}{4}} \right)\\
= \frac{{ - 145\sqrt {33} }}{{16}}
\end{array}$
