Tìm x,y,z biết: (x-y-7)^2+(4x-3y-24)^2=0 26/07/2021 Bởi Kaylee 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 \(x = 3;\,\,y = – 4\) Bình luận
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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 \(x = 3;\,\,y = – 4\)