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Ta có: \(\left\{{}\begin{matrix}\left(2x-3y\right)^{2018}\ge0\forall x,y\\\left(3y-4z\right)^{2020}\ge0\forall y,z\\\left|2x+3y-z-63\right|\ge0\forall x,y,z\end{matrix}\right.\)
\(\Rightarrow\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|\ge0\forall x,y,z\)
Mà: \(\left(2x-3y\right)^{2018}+\left(3y-4z\right)^{2020}+\left|2x+3y-z-63\right|=0\)
nên: \(\left\{{}\begin{matrix}2x-3y=0\\3y-4z=0\\2x+3y-z-63=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x=3y\\3y=4z\\z=2x+3y-63\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=4z\\3y=4z\\z=4z+4z-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4z:2\\y=4z:3\\z=8z-63\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=4z:3\\-7z=-63\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=4\cdot9:3=12\\z=9\end{matrix}\right.\)
Vậy \(x=18;y=12;z=9\).
$Toru$
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Vì: |2\(x\) - 1| = |1 - 2\(x\)|
Nên: |2\(x\) - 1| + |1 - 2\(x\)| = 8
⇒ |2\(x\) - 1| + |2\(x\) - 1| = 8
2.|2\(x\) - 1| = 8
|2\(x\) - 1| = 8:2
|2\(x\) - 1| = 4
\(\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4+1\\2x=4+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-3\\2x=5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\){- \(\dfrac{3}{2}\); \(\dfrac{5}{2}\)}
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