Tìm x,y thuộc Z biết 3x^2-y^2-2xy-2x-2y+40=0 19/07/2021 Bởi Liliana Tìm x,y thuộc Z biết 3x^2-y^2-2xy-2x-2y+40=0
Đáp án: \[\left[ \begin{array}{l}x = y = 10\\\left\{ \begin{array}{l}x = – 10\\y = – 12\end{array} \right.\\\left\{ \begin{array}{l}x = – 10\\y = 30\end{array} \right.\\\left\{ \begin{array}{l}x = 10\\y = – 32\end{array} \right.\end{array} \right.\] Giải thích các bước giải: \(\begin{array}{l}3{x^2} – {y^2} – 2xy – 2x – 2y + 40 = 0\\ \Leftrightarrow 4{x^2} – \left( {{x^2} + {y^2} + 1 + 2xy + 2x + 2y} \right) + 41 = 0\\ \Leftrightarrow {\left( {2x} \right)^2} – {\left( {x + y + 1} \right)^2} = – 41\\ \Leftrightarrow {\left( {x + y + 1} \right)^2} – {\left( {2x} \right)^2} = 41\\ \Leftrightarrow \left( {3x + y + 1} \right)\left( { – x + y + 1} \right) = 41\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}3x + y + 1 = 41\\ – x + y + 1 = 1\end{array} \right.\\\left\{ \begin{array}{l}3x + y + 1 = – 41\\ – x + y + 1 = – 1\end{array} \right.\\\left\{ \begin{array}{l}3x + y + 1 = 1\\ – x + y + 1 = 41\end{array} \right.\\\left\{ \begin{array}{l}3x + y + 1 = – 1\\ – x + y + 1 = – 41\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = y = 10\\\left\{ \begin{array}{l}x = – 10\\y = – 12\end{array} \right.\\\left\{ \begin{array}{l}x = – 10\\y = 30\end{array} \right.\\\left\{ \begin{array}{l}x = 10\\y = – 32\end{array} \right.\end{array} \right.\end{array}\) Bình luận
Đáp án:
\[\left[ \begin{array}{l}
x = y = 10\\
\left\{ \begin{array}{l}
x = – 10\\
y = – 12
\end{array} \right.\\
\left\{ \begin{array}{l}
x = – 10\\
y = 30
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 10\\
y = – 32
\end{array} \right.
\end{array} \right.\]
Giải thích các bước giải:
\(\begin{array}{l}
3{x^2} – {y^2} – 2xy – 2x – 2y + 40 = 0\\
\Leftrightarrow 4{x^2} – \left( {{x^2} + {y^2} + 1 + 2xy + 2x + 2y} \right) + 41 = 0\\
\Leftrightarrow {\left( {2x} \right)^2} – {\left( {x + y + 1} \right)^2} = – 41\\
\Leftrightarrow {\left( {x + y + 1} \right)^2} – {\left( {2x} \right)^2} = 41\\
\Leftrightarrow \left( {3x + y + 1} \right)\left( { – x + y + 1} \right) = 41\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x + y + 1 = 41\\
– x + y + 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + y + 1 = – 41\\
– x + y + 1 = – 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + y + 1 = 1\\
– x + y + 1 = 41
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + y + 1 = – 1\\
– x + y + 1 = – 41
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = y = 10\\
\left\{ \begin{array}{l}
x = – 10\\
y = – 12
\end{array} \right.\\
\left\{ \begin{array}{l}
x = – 10\\
y = 30
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 10\\
y = – 32
\end{array} \right.
\end{array} \right.
\end{array}\)