Toán Lớp 9: Gi ải hệ pt: $\left \{ {{x+y+xy+1=0 } \atop {x^{2}+y^{2}-x-y=22}} \right.$
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Toán Lớp 9: Gi ải hệ pt: $\left \{ {{x+y+xy+1=0 } \atop {x^{2}+y^{2}-x-y=22}} \right.$
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\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}\)