Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
a + b = 5\\
a.b = - 2
\end{array} \right.\\
+ ){a^2} + {b^2}\\
= {a^2} + 2ab + {b^2} - 2ab\\
= {\left( {a + b} \right)^2} - 2ab\\
= {5^2} - 2.\left( { - 2} \right)\\
= 25 + 4 = 29\\
+ )\dfrac{1}{{{a^3}}} + \dfrac{1}{{{b^3}}}\\
= \dfrac{{{b^3} + {a^3}}}{{{a^3}{b^3}}}\\
= \dfrac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{{{\left( {ab} \right)}^3}}}\\
= \dfrac{{5.\left[ {{{\left( {a + b} \right)}^2} - 3ab} \right]}}{{{{\left( { - 2} \right)}^3}}}\\
= - \dfrac{{5\left( {{5^2} - 3.\left( { - 2} \right)} \right)}}{8}\\
= - \dfrac{{5.\left( {25 + 6} \right)}}{8}\\
= - \dfrac{{155}}{8}\\
+ )a - b\\
= - \sqrt {{{\left( {a - b} \right)}^2}} \left( {do:a < b} \right)\\
= - \sqrt {{{\left( {a + b} \right)}^2} - 4ab} \\
= - \sqrt {{5^2} - 4.\left( { - 2} \right)} \\
= - \sqrt {33} \\
+ ){a^3} - {b^3}\\
= \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= - \sqrt {33} .\left[ {{{\left( {a + b} \right)}^2} - ab} \right]\\
= - \sqrt {33} .\left( {{5^2} - \left( { - 2} \right)} \right)\\
= - \sqrt {33} .27\\
= - 27\sqrt {33}
\end{array}$
