Đáp án:
$\left( {x;\,\,y} \right) \in \left\{ {\left( {3; - 1} \right);\,\,\left( {3; - 3} \right);\,\,\left( { - 1; - 1} \right);\,\,\left( { - 1; - 3} \right)} \right\}.$
Giải thích các bước giải:
\(\begin{array}{l}\left\{ \begin{array}{l}5\left| {x - 1} \right| - 3\left| {y + 2} \right| = 7\\2\sqrt {4{x^2} - 8x + 4} + 5\sqrt {{y^2} + 4y + 4} = 13\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}5\left| {x - 1} \right| - 3\left| {y + 2} \right| = 7\\2\sqrt {4{{\left( {x - 1} \right)}^2}} + 5\sqrt {{{\left( {y + 2} \right)}^2}} = 13\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}5\left| {x - 1} \right| - 3\left| {y + 2} \right| = 7\\4\left| {x - 1} \right| + 5\left| {y + 2} \right| = 13\end{array} \right.\end{array}\)
Đặt \(\left\{ \begin{array}{l}\left| {x - 1} \right| = a\,\,\left( {a \ge 0} \right)\\\left| {y + 2} \right| = b\,\,\,\left( {b \ge 0} \right)\end{array} \right..\) Khi đó ta có hệ phương trình:
\(\begin{array}{l}\left\{ \begin{array}{l}5a - 3b = 7\\4a + 5b = 13\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}25a - 15b = 35\\12a + 15b = 39\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}37a = 74\\5a - 3b = 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2\,\,\,\left( {tm} \right)\\b = 1\,\,\,\,\left( {tm} \right)\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}\left| {x - 1} \right| = 2\\\left| {y + 2} \right| = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x - 1 = 2\\x - 1 = - 2\end{array} \right.\\\left[ \begin{array}{l}y + 2 = 1\\y + 2 = - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 3\\x = - 1\end{array} \right.\\\left[ \begin{array}{l}y = - 1\\y = - 3\end{array} \right.\end{array} \right..\end{array}\)
Vậy hệ phương trình đã cho có nghiệm: \(\left( {x;\,\,y} \right) \in \left\{ {\left( {3; - 1} \right);\,\,\left( {3; - 3} \right);\,\,\left( { - 1; - 1} \right);\,\,\left( { - 1; - 3} \right)} \right\}.\)
