Tìm x,y,z biết: (x-y-7)^2+(4x-3y-24)^2=0

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 \(x = 3;\,\,y =  – 4\)