Toán Lớp 9: tìm nghiệm nguyên của pt: x²-4x+2y-xy+9=0
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Toán Lớp 9: tìm nghiệm nguyên của pt: x²-4x+2y-xy+9=0
Giúp mình vs ạ
Comments ( 1 )
{x^2} – 4x + 2y – xy + 9 = 0\\
\Leftrightarrow {x^2} – 4x + 4 + y\left( {2 – x} \right) + 5 = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} – y\left( {x – 2} \right) = – 5\\
\Leftrightarrow \left( {x – 2} \right)\left( {x – 2 – y} \right) = – 5 = – 1.5 = \left( { – 5} \right).1\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 = – 1\\
x – 2 – y = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = 1\\
x – 2 – y = – 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = 5\\
x – 2 – y = – 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 = – 5\\
x – 2 – y = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 1\\
y = x – 7 = – 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 3\\
y = x + 3 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 7\\
y = x – 1 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = – 3\\
y = x – 3 = – 6
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1; – 6} \right);\left( {3;6} \right);\left( {7;6} \right);\left( { – 3; – 6} \right)} \right\}
\end{array}$