Đáp án:
\[\left( {x;y} \right) = \left\{ {\left( {4;1} \right);\left( {1;4} \right)} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^3} + {y^3} = 65\,\,\,\,\,\left( 1 \right)\\
{x^2}y + {y^2}x = 20
\end{array} \right.\\
\left( {{x^3} + {y^3}} \right) + 3\left( {{x^2}y + {y^2}x} \right) = 65 + 3.20\\
\Leftrightarrow {x^3} + 3{x^2}y + 3x{y^2} + {y^3} = 125\\
\Leftrightarrow {\left( {x + y} \right)^3} = {5^3}\\
\Leftrightarrow x + y = 5\\
\Leftrightarrow y = 5 - x\\
\left( 1 \right) \Leftrightarrow {x^3} + {\left( {5 - x} \right)^3} = 65\\
\Leftrightarrow {x^3} + 125 - 75x + 15{x^2} - {x^3} = 65\\
\Leftrightarrow 15{x^2} - 75x + 60 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4 \Rightarrow y = 1\\
x = 1 \Rightarrow y = 4
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {4;1} \right);\left( {1;4} \right)} \right\}
\end{array}\)
