a, \(x\left(x-\dfrac{1}{2}\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-\dfrac{1}{2}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-\dfrac{1}{2}>0\end{matrix}\right.\end{matrix}\right.\Rightarrow0< x< \dfrac{1}{2}\)
Vậy \(0< x< \dfrac{1}{2}\)
b, \(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|=0\)
Mà \(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{4}-x\right|=0\\\left|x-y+z\right|=0\\\left|\dfrac{2}{3}+y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-11}{12}\\z=\dfrac{-2}{3}\end{matrix}\right.\)
Vậy...
Sửa đề: \(x;y;z\in Q\)
\(x\left(x-\dfrac{1}{2}\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-\dfrac{1}{2}< 0\Rightarrow x< \dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-\dfrac{1}{2}>0\Rightarrow x>\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow0< x< \dfrac{1}{2}\)
\(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|=0\)
\(\left\{{}\begin{matrix}\left|\dfrac{1}{4}-x\right|\ge0\\\left|x-y+z\right|\ge0\\\left|\dfrac{2}{3}+y\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\)\(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{1}{4}-x\right|=0\Rightarrow x=\dfrac{1}{4}\\\left|\dfrac{2}{3}+y\right|=0\Rightarrow y=-\dfrac{2}{3}\\\left|x-y+z\right|=0\Rightarrow z=-\dfrac{11}{12}\end{matrix}\right.\)
a, \(x\left(x-\dfrac{1}{2}\right)< 0\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-\dfrac{1}{2}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-\dfrac{1}{2}>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\dfrac{1}{2}< x< 0\)
Vậy...
b, \(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|=0\)
Mà \(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{4}-x=0\\x-y+z=0\\\dfrac{2}{3}+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\z=\dfrac{-11}{12}\\y=\dfrac{-2}{3}\end{matrix}\right.\)
Vậy...
