Đáp án:
\[\left( {x;y} \right) = \left\{ {\left( { - 1;5} \right);\left( { - 1; - 4} \right);\left( { - 4; - 1} \right);\left( {5; - 1} \right)} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + y + xy + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
{x^2} + {y^2} - x - y = 22\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {x + xy} \right) + \left( {y + 1} \right) = 0\\
\Leftrightarrow x.\left( {1 + y} \right) + \left( {y + 1} \right) = 0\\
\Leftrightarrow \left( {y + 1} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
y + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
y = - 1
\end{array} \right.\\
TH1:\,\,\,\,x = - 1\\
\left( 2 \right) \Leftrightarrow {\left( { - 1} \right)^2} + {y^2} - \left( { - 1} \right) - y = 22\\
\Leftrightarrow 1 + {y^2} + 1 - y - 22 = 0\\
\Leftrightarrow {y^2} - y - 20 = 0\\
\Leftrightarrow \left( {{y^2} - 5y} \right) + \left( {4y - 20} \right) = 0\\
\Leftrightarrow y\left( {y - 5} \right) + 4\left( {y - 5} \right) = 0\\
\Leftrightarrow \left( {y - 5} \right)\left( {y + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y - 5 = 0\\
y + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 5\\
y = - 4
\end{array} \right.\\
TH2:\,\,\,\,y = - 1\\
\left( 2 \right) \Leftrightarrow {x^2} + {\left( { - 1} \right)^2} - x - \left( { - 1} \right) = 22\\
\Leftrightarrow {x^2} + 1 - x + 1 - 22 = 0\\
\Leftrightarrow {x^2} - x - 20 = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = - 4
\end{array} \right.
\end{array}\)
Vậy cặp nghiệm của hệ phương trình đã cho là:
\(\left( {x;y} \right) = \left\{ {\left( { - 1;5} \right);\left( { - 1; - 4} \right);\left( { - 4; - 1} \right);\left( {5; - 1} \right)} \right\}\)
