Đáp án:
\[x = 3;\,\,y = - 4\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {x - y - 7} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {4x - 3y - 24} \right)^2} \ge 0,\,\,\,\forall x,y\\
\Rightarrow {\left( {x - y - 7} \right)^2} + {\left( {4x - 3y - 24} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x - y - 7} \right)^2} + {\left( {4x - 3y - 24} \right)^2} = 0\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x - y - 7} \right)^2} = 0\\
{\left( {4x - 3y - 24} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y - 7 = 0\\
4x - 3y - 24 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y + 7\\
4x - 3y - 24 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y + 7\\
4.\left( {y + 7} \right) - 3y - 24 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y + 7\\
4y + 28 - 3y - 24 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y + 7\\
y + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y + 7\\
y = - 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = - 4
\end{array} \right.
\end{array}\)
Vậy \(x = 3;\,\,y = - 4\)
