$\begin{array}{l}Xe't:\left\{ \begin{array}{l}x - 2y + 1 = 0\\2x + ay + 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 2y = - 1\\2x + ay = - 5\end{array} \right.\, \Rightarrow D = \left| \begin{array}{l}1\,\,\, - 2\\2\,\,\,\,\,\,\,a\end{array} \right| = a + 4\\ \bullet \,Khi\,a \ne - 4 \Rightarrow \left\{ \begin{array}{l}{D_x} = \left| \begin{array}{l} - 1\,\,\,\,\, - 2\\ - 5\,\,\,\,\,\,\,a\end{array} \right| = - \left( {a + 10} \right) \Rightarrow x = - \frac{{a + 10}}{{a + 4}}\\{D_y} = \left| \begin{array}{l}1\,\,\,\,\,\, - 1\\2\,\,\,\,\,\, - 5\end{array} \right| = - 3 \Rightarrow y = - \frac{3}{{a + 4}}\end{array} \right.\\Do\,\left\{ \begin{array}{l}{\left( {x - 2y + 1} \right)^2} \ge 0\\{\left( {2x + ay + 5} \right)^2} \ge 0\end{array} \right. \Rightarrow T \ge 0 \Rightarrow Min\,T = 0 \Leftrightarrow \left( {x;y} \right) = \left( { - \frac{{a + 10}}{{a + 4}}; - \frac{3}{{a + 4}}} \right)\\ \bullet \,Khi\,a = - 4 \Rightarrow T = {\left( {x - 2y + 1} \right)^2} + {\left( {2x - 4y + 5} \right)^2} = {\left( {z + 1} \right)^2} + {\left( {2z + 5} \right)^2}\left( {z = x - 2y} \right)\\ \Rightarrow T = 5{z^2} + 22z + 26 = 5\left( {{z^2} + \frac{{22}}{5}z + \frac{{121}}{{25}}} \right) + 26 - \frac{{121}}{5} = 5{\left( {z + \frac{{11}}{5}} \right)^2} + \frac{9}{5} \ge \frac{9}{5}\\ \Rightarrow Min\,T = \frac{9}{5} \Leftrightarrow z = - \frac{{11}}{5} = x - 2y \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {{x_0};{y_0}} \right);5{x_0} - 10{y_0} + 11 = 0} \right\}\end{array}$
\[\begin{array}{l}Xe't:\left\{ \begin{array}{l}x - 2y + 1 = 0\\2x + ay + 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 2y = - 1\\2x + ay = - 5\end{array} \right.\, \Rightarrow D = \left| \begin{array}{l}1\,\,\, - 2\\2\,\,\,\,\,\,\,a\end{array} \right| = a + 4\\ \bullet \,Khi\,a \ne - 4 \Rightarrow \left\{ \begin{array}{l}{D_x} = \left| \begin{array}{l} - 1\,\,\,\,\, - 2\\ - 5\,\,\,\,\,\,\,a\end{array} \right| = - \left( {a + 10} \right) \Rightarrow x = - \dfrac{{a + 10}}{{a + 4}}\\{D_y} = \left| \begin{array}{l}1\,\,\,\,\,\, - 1\\2\,\,\,\,\,\, - 5\end{array} \right| = - 3 \Rightarrow y = - \dfrac{3}{{a + 4}}\end{array} \right.\\Do\,\left\{ \begin{array}{l}{\left( {x - 2y + 1} \right)^2} \ge 0\\{\left( {2x + ay + 5} \right)^2} \ge 0\end{array} \right. \Rightarrow T \ge 0 \Rightarrow Min\,T = 0 \Leftrightarrow \left( {x;y} \right) = \left( { - \dfrac{{a + 10}}{{a + 4}}; - \dfrac{3}{{a + 4}}} \right)\\ \bullet \,Khi\,a = - 4 \Rightarrow T = {\left( {x - 2y + 1} \right)^2} + {\left( {2x - 4y + 5} \right)^2} = {\left( {z + 1} \right)^2} + {\left( {2z + 5} \right)^2}\left( {z = x - 2y} \right)\\ \Rightarrow T = 5{z^2} + 22z + 26 = 5\left( {{z^2} + \dfrac{{22}}{5}z + \dfrac{{121}}{{25}}} \right) + 26 - \dfrac{{121}}{5} = 5{\left( {z + \dfrac{{11}}{5}} \right)^2} + \dfrac{9}{5} \ge \dfrac{9}{5}\\ \Rightarrow Min\,T = \dfrac{9}{5} \Leftrightarrow z = - \dfrac{{11}}{5} = x - 2y \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {{x_0};{y_0}} \right);5{x_0} - 10{y_0} + 11 = 0} \right\}\end{array}\]
$\begin{array}{l}Xe't:\left\{ \begin{array}{l}x - 2y + 1 = 0\\2x + ay + 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 2y = - 1\\2x + ay = - 5\end{array} \right.\, \Rightarrow D = \left| \begin{array}{l}1\,\,\, - 2\\2\,\,\,\,\,\,\,a\end{array} \right| = a + 4\\ \bullet \,Khi\,a \ne - 4 \Rightarrow \left\{ \begin{array}{l}{D_x} = \left| \begin{array}{l} - 1\,\,\,\,\, - 2\\ - 5\,\,\,\,\,\,\,a\end{array} \right| = - \left( {a + 10} \right) \Rightarrow x = - \frac{{a + 10}}{{a + 4}}\\{D_y} = \left| \begin{array}{l}1\,\,\,\,\,\, - 1\\2\,\,\,\,\,\, - 5\end{array} \right| = - 3 \Rightarrow y = - \frac{3}{{a + 4}}\end{array} \right.\\Do\,\left\{ \begin{array}{l}{\left( {x - 2y + 1} \right)^2} \ge 0\\{\left( {2x + ay + 5} \right)^2} \ge 0\end{array} \right. \Rightarrow T \ge 0 \Rightarrow Min\,T = 0 \Leftrightarrow \left( {x;y} \right) = \left( { - \frac{{a + 10}}{{a + 4}}; - \frac{3}{{a + 4}}} \right)\\ \bullet \,Khi\,a = - 4 \Rightarrow T = {\left( {x - 2y + 1} \right)^2} + {\left( {2x - 4y + 5} \right)^2} = {\left( {z + 1} \right)^2} + {\left( {2z + 5} \right)^2}\left( {z = x - 2y} \right)\\ \Rightarrow T = 5{z^2} + 22z + 26 = 5\left( {{z^2} + \frac{{22}}{5}z + \frac{{121}}{{25}}} \right) + 26 - \frac{{121}}{5} = 5{\left( {z + \frac{{11}}{5}} \right)^2} + \frac{9}{5} \ge \frac{9}{5}\\ \Rightarrow Min\,T = \frac{9}{5} \Leftrightarrow z = - \frac{{11}}{5} = x - 2y \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {{x_0};{y_0}} \right);5{x_0} - 10{y_0} + 11 = 0} \right\}\end{array} $
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