Đáp án:
\[\left[ \begin{array}{l}
x = 1;\,\,\,y = 2\\
x = 2;\,\,\,\,y = 1
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + y + xy = 5\\
{\left( {x + 1} \right)^3} + {\left( {y + 1} \right)^3} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right) + \left( {y + xy} \right) = 6\\
{\left( {x + 1} \right)^3} + {\left( {y + 1} \right)^3} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right) + y\left( {x + 1} \right) = 6\\
{\left( {x + 1} \right)^3} + {\left( {y + 1} \right)^3} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {y + 1} \right) = 6\\
{\left( {x + 1} \right)^3} + {\left( {y + 1} \right)^3} = 35
\end{array} \right.
\end{array}\)
Đặt \(a = x + 1;\,\,\,b = y + 1\), khi đó, hệ phương trình trên trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
ab = 6\\
{a^3} + {b^3} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{6}{a}\\
{a^3} + {\left( {\dfrac{6}{a}} \right)^3} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{6}{a}\\
{a^3} + \dfrac{{216}}{{{a^3}}} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{6}{a}\\
{a^6} - 35{a^3} + 216 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{6}{a}\\
\left( {{a^6} - 27{a^3}} \right) - \left( {8{a^3} - 216} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{6}{a}\\
{a^3}\left( {{a^3} - 27} \right) - 8.\left( {{a^3} - 27} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{6}{a}\\
\left( {{a^3} - 8} \right)\left( {{a^3} - 27} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = 2;\,\,\,b = 3\\
a = 3;\,\,a = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1;\,\,\,y = 2\\
x = 2;\,\,\,\,y = 1
\end{array} \right.
\end{array}\)
