\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|x-\frac{1}{2}\right|\ge0\\\left|y+\frac{2}{3}\right|\ge0\\\left|x^2+xz\right|\ge0\end{matrix}\right.\forall x,y,z.\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|\ge0\) \(\forall x,y,z.\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)
Ta có 3 trường hợp.
+ TH1: \(\left|x-\frac{1}{2}\right|=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=0+\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\)
+ TH2: \(\left|y+\frac{2}{3}\right|=0\)
\(\Rightarrow y+\frac{2}{3}=0\)
\(\Rightarrow y=0-\frac{2}{3}\)
\(\Rightarrow y=-\frac{2}{3}.\)
+ TH3: \(\left|x^2+xz\right|=0\)
\(\Rightarrow x^2+xz=0\)
\(\Rightarrow\left(\frac{1}{2}\right)^2+\frac{1}{2}z=0\)
\(\Rightarrow\frac{1}{4}+\frac{1}{2}z=0\)
\(\Rightarrow\frac{1}{2}z=0-\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}z=-\frac{1}{4}\)
\(\Rightarrow z=\left(-\frac{1}{4}\right):\frac{1}{2}\)
\(\Rightarrow z=-\frac{1}{2}.\)
Vậy \(\left(x;y;z\right)\in\left\{\frac{1}{2};-\frac{2}{3};-\frac{1}{2}\right\}.\)
Chúc bạn học tốt!
