Tìm x,y thuộc Z biết 3x^2-y^2-2xy-2x-2y+40=0

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Tìm x,y thuộc Z biết 3x^2-y^2-2xy-2x-2y+40=0

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2 tháng 2021-07-19T11:01:17+00:00 1 Answers 0 views 0

$\left[ \begin{array}{l} x = y = 10\\ \left\{ \begin{array}{l} x = – 10\\ y = – 12 \end{array} \right.\\ \left\{ \begin{array}{l} x = – 10\\ y = 30 \end{array} \right.\\ \left\{ \begin{array}{l} x = 10\\ y = – 32 \end{array} \right. \end{array} \right.$
$$\begin{array}{l} 3{x^2} – {y^2} – 2xy – 2x – 2y + 40 = 0\\ \Leftrightarrow 4{x^2} – \left( {{x^2} + {y^2} + 1 + 2xy + 2x + 2y} \right) + 41 = 0\\ \Leftrightarrow {\left( {2x} \right)^2} – {\left( {x + y + 1} \right)^2} = – 41\\ \Leftrightarrow {\left( {x + y + 1} \right)^2} – {\left( {2x} \right)^2} = 41\\ \Leftrightarrow \left( {3x + y + 1} \right)\left( { – x + y + 1} \right) = 41\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 3x + y + 1 = 41\\ – x + y + 1 = 1 \end{array} \right.\\ \left\{ \begin{array}{l} 3x + y + 1 = – 41\\ – x + y + 1 = – 1 \end{array} \right.\\ \left\{ \begin{array}{l} 3x + y + 1 = 1\\ – x + y + 1 = 41 \end{array} \right.\\ \left\{ \begin{array}{l} 3x + y + 1 = – 1\\ – x + y + 1 = – 41 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = y = 10\\ \left\{ \begin{array}{l} x = – 10\\ y = – 12 \end{array} \right.\\ \left\{ \begin{array}{l} x = – 10\\ y = 30 \end{array} \right.\\ \left\{ \begin{array}{l} x = 10\\ y = – 32 \end{array} \right. \end{array} \right. \end{array}$$