Đáp án:
$P = \dfrac{{12555}}{4}$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + {y^2} = {\left( {x + y} \right)^2} - 2xy = {5^2} - 2\left( { - 2} \right) = 29\\
{x^4} + {y^4} = {\left( {{x^2} + {y^2}} \right)^2} - 2{x^2}{y^2} = {29^2} - 2{\left( { - 2} \right)^2} = 833
\end{array} \right.\\
\Rightarrow \dfrac{{{x^3}}}{{{y^2}}} + \dfrac{{{y^3}}}{{{x^2}}} = \dfrac{{{x^5} + {y^5}}}{{{x^2}{y^2}}} = \dfrac{{\left( {x + y} \right)\left( {{x^4} - {x^3}y + {x^2}{y^2} - x{y^3} + {y^4}} \right)}}{{{{\left( {xy} \right)}^2}}}\\
= \dfrac{{\left( {x + y} \right)\left[ {\left( {{x^4} + {y^4}} \right) - xy\left( {{x^2} + {y^2}} \right) + {{\left( {xy} \right)}^2}} \right]}}{{{{\left( {xy} \right)}^2}}}\\
= \dfrac{{5\left( {833 - \left( { - 2} \right).29 + {{\left( { - 2} \right)}^2}} \right)}}{{{{\left( { - 2} \right)}^2}}} = \dfrac{{4475}}{4}\\
\Rightarrow P = \dfrac{{{x^3}}}{{{y^2}}} + \dfrac{{{y^3}}}{{{x^2}}} + 2020 = \dfrac{{12555}}{4}
\end{array}$
Vậy $P = \dfrac{{12555}}{4}$
