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\(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|=0\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|y+\dfrac{2}{3}\right|\ge0\forall y\\\left|x^2+xz\right|\ge0\forall x;z\end{matrix}\right.\) \(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|y+\dfrac{2}{3}\right|=0\\\left|x^2+xz\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{2}{3}\\z=-\dfrac{1}{2}\end{matrix}\right.\)
Ta có: \(\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\left|y+\dfrac{2}{3}\right|\ge0\forall x\)
\(\left|x^2+xz\right|\ge0\forall x,z\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|\ge0\forall x,y,z\)
Để \(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\y+\dfrac{2}{3}=0\Rightarrow y=\dfrac{-2}{3}\\x^2+xz=0\Rightarrow x\left(x+z\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\z=0,x=0\end{matrix}\right.\end{matrix}\right.\)
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