cho x,y là các số thực thỏa mãn x^2 + 2y^2 + 2xy + 2x + 6y +1 = 0 16/11/2021 Bởi Arya cho x,y là các số thực thỏa mãn x^2 + 2y^2 + 2xy + 2x + 6y +1 = 0
Đáp án: $\begin{array}{l}{x^2} + 2{y^2} + 2xy + 2x + 6y + 1 = 0\\ \Rightarrow {x^2} + {y^2} + 1 + 2x + 2y + 2xy\\ + {y^2} + 4y + 4 = 4\\ \Rightarrow {\left( {x + y + 1} \right)^2} + {\left( {y + 2} \right)^2} = 4 = 0 + 4\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{\left( {x + y + 1} \right)^2} = 4\\{\left( {y + 2} \right)^2} = 0\end{array} \right.\\\left\{ \begin{array}{l}{\left( {x + y + 1} \right)^2} = 0\\{\left( {y + 2} \right)^2} = 4\end{array} \right.\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + y + 1 = 2/x + y + 1 = – 2\\y = – 2\end{array} \right.\\\left\{ \begin{array}{l}x + y + 1 = 0\\\left[ \begin{array}{l}y = 0\\y = – 4\end{array} \right.\end{array} \right.\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = 3;y = – 2\\x = – 1;y = – 2\\x = – 1;y = 0\\x = 3;y = – 4\end{array} \right.\\Vậy\,\left( {x;y} \right) = \left\{ {\left( {3; – 2} \right);\left( { – 1; – 2} \right);\left( { – 1;0} \right);\left( {3; – 4} \right)} \right\}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
{x^2} + 2{y^2} + 2xy + 2x + 6y + 1 = 0\\
\Rightarrow {x^2} + {y^2} + 1 + 2x + 2y + 2xy\\
+ {y^2} + 4y + 4 = 4\\
\Rightarrow {\left( {x + y + 1} \right)^2} + {\left( {y + 2} \right)^2} = 4 = 0 + 4\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + y + 1} \right)^2} = 4\\
{\left( {y + 2} \right)^2} = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {x + y + 1} \right)^2} = 0\\
{\left( {y + 2} \right)^2} = 4
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + y + 1 = 2/x + y + 1 = – 2\\
y = – 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y + 1 = 0\\
\left[ \begin{array}{l}
y = 0\\
y = – 4
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3;y = – 2\\
x = – 1;y = – 2\\
x = – 1;y = 0\\
x = 3;y = – 4
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {3; – 2} \right);\left( { – 1; – 2} \right);\left( { – 1;0} \right);\left( {3; – 4} \right)} \right\}
\end{array}$