Đáp án:
x=2,y=1
x=1,y=2
Giải thích các bước giải:
\(\left\{ \begin{array}{l}
{x^3} + {x^3}{y^3} + {y^3} = 17\\
x + y + xy = 5
\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}
x + y = a\\
xy = b
\end{array} \right.\)
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
{a^3} - 3ab + {b^3} = 17\\
a + b = 5
\end{array} \right.\\
{a^3} - 3ab + {b^3} = 17 \leftrightarrow {(a + b)^3} - 3ab(a + b) - 3ab = 17\\
\leftrightarrow {5^3} - 3ab.5 - 3ab = 17\\
\leftrightarrow ab = 6\\
\to \left\{ \begin{array}{l}
a + b = 5\\
ab = 6
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 5 - b\\
b(5 - b) = 6
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 5 - b\\
- {b^2} + 5b - 6 = 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
a = 5 - b\\
\left[ \begin{array}{l}
b = 3\\
b = 2
\end{array} \right.
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
a = 2,b = 3\\
a = 3,b = 2
\end{array} \right.
\end{array}\)
Xét a=2,b=3
\( \to \left\{ \begin{array}{l}
x + y = 2\\
xy = 3
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 2 - y\\
y(2 - y) = 3
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 2 - y\\
- {y^2} + 2y = 3(VN)
\end{array} \right. \to x,y \in \phi \)
Xét a=3,b=2
\( \to \left\{ \begin{array}{l}
x + y = 3\\
xy = 2
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 3 - y\\
y(3 - y) = 2
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = 3 - y\\
- {y^2} + 3y = 2(VN)
\end{array} \right. \to \left\{ \begin{array}{l}
x = 3 - y\\
\left[ \begin{array}{l}
y = 1\\
y = 2
\end{array} \right.
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = 2,y = 1\\
x = 1,y = 2
\end{array} \right.\)
