(x-y-7 mũ 2 ) + (4x -3y -24) mũ 2 = 0 tìm x,y

Đáp án:

$\begin{array}{l}
{\left( {x – y – 7} \right)^2} + {\left( {4x – 3y – 24} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x – y – 7} \right)^2} \ge 0\\
{\left( {4x – 3y – 24} \right)^2} \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {x – y – 7} \right)^2} = 0\\
{\left( {4x – 3y – 24} \right)^2} = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x – y – 7 = 0\\
4x – 3y – 24 = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = y + 7\\
4x – 3y – 24 = 0
\end{array} \right.\\
Thay\,x = y + 7\,vao\,4x – 3y – 24 = 0\\
 \Rightarrow 4.\left( {y + 7} \right) – 3y – 24 = 0\\
 \Rightarrow 4y + 28 – 3y – 24 = 0\\
 \Rightarrow y + 4 = 0\\
 \Rightarrow y =  – 4\\
 \Rightarrow x = y + 7 =  – 4 + 7 = 3\\
Vậy\,x = 3;y =  – 4
\end{array}$