Với mọi giá trị của x;y;z ta có:
\(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|\ge0\)
Để \(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|=0\\\left|y-\dfrac{1}{4}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\\dfrac{1}{4}+\dfrac{1}{4}+z=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\z=-\dfrac{1}{4}-\dfrac{1}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..........
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Ta có: \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|\ge0\\\left|y-\dfrac{1}{4}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\Rightarrow\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|\ge0\)
Mà \(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|=0\\\left|y-\dfrac{1}{4}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\z=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\dfrac{1}{4};\dfrac{1}{4};\dfrac{-1}{2}\right)\)
Ta có: \(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\)
Mà \(\left|x-\dfrac{1}{4}\right|\ge0;\left|y-\dfrac{1}{4}\right|\ge0;\left|x+y+z\right|\ge0\)
Để \(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\) thì
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\y-\dfrac{1}{4}=0\\x+y+z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\z=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\z=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|\ge0\\\left|y-\dfrac{1}{4}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|=0\Rightarrow x=\dfrac{1}{4}\\\left|y-\dfrac{1}{4}\right|=0\Rightarrow y=\dfrac{1}{4}\\\left|x+y+z\right|=0\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)
Do
\(\left|x-\dfrac{1}{4}\right|\ge0\forall x\)
\(\left|y-\dfrac{1}{4}\right|\ge0\forall y\)
\(\left|x+y+z\right|\ge0\forall x,y,z\)
\(\Rightarrow\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)
Để \(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\)
=>\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|=0\\\left|y-\dfrac{1}{4}\right|=0\\\left|z+x+y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\\left|x+y+z\right|=0\end{matrix}\right.\)
Thay \(x=\dfrac{1}{4};y=\dfrac{1}{4}\) và |x+y+z|=0 ta tính ra đc \(z=-\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{4};y=\dfrac{1}{4}\);\(z=-\dfrac{1}{2}\)
