Đáp án:
\(\left[ \begin{array}{l}
{x^3} + {y^3} = 8\sqrt 5 \\
{x^3} + {y^3} = - 8\sqrt 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Do:x - y = 3\\
\to x = y + 3\\
Thay:x = y + 3\\
\to {\left( {y + 3} \right)^2} + {y^2} = 7\\
\to 2{y^2} + 6y + 9 = 7\\
\to 2{y^2} + 6y + 2 = 0\\
\to {y^2} + 3y + 1 = 0\\
\to {y^2} + 2.y.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{5}{4} = 0\\
\to {\left( {y + \dfrac{3}{2}} \right)^2} = \dfrac{5}{4}\\
\to \left| {y + \dfrac{3}{2}} \right| = \sqrt {\dfrac{5}{4}} \\
\to \left[ \begin{array}{l}
y + \dfrac{3}{2} = \dfrac{{\sqrt 5 }}{2}\\
y + \dfrac{3}{2} = - \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = \dfrac{{ - 3 + \sqrt 5 }}{2}\\
y = \dfrac{{ - 3 - \sqrt 5 }}{2}
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt 5 }}{2}\\
x = \dfrac{{3 - \sqrt 5 }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^3} + {y^3} = 8\sqrt 5 \\
{x^3} + {y^3} = - 8\sqrt 5
\end{array} \right.
\end{array}\)
