\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\Rightarrow x=\dfrac{1}{2}\)
Giải:
\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\)
Vì \(\left|x-\dfrac{1}{2}\right|\ge0;\forall x\) và \(\left|x+y\right|\ge0;\forall x,y\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{2}\) và \(y=-\dfrac{1}{2}\).
Chúc bạn học tốt!
\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0+\dfrac{1}{2}=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=0-\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{2};y=-\dfrac{1}{2}\)
