Question

\(\left|x+y\right|+\left|y-\frac{1}{2}\right|=0\)

Answer 1

Có: \(\left|x+y\right|\ge0;\left|y-\frac{1}{2}\right|\ge0\forall x;y\)

Mà theo đề bài: \(\left|x+y\right|+\left|y-\frac{1}{2}\right|=0\)

\(\Rightarrow\begin{cases}\left|x+y\right|=0\\\left|y-\frac{1}{2}\right|=0\end{cases}\)\(\Rightarrow\begin{cases}x+y=0\\y-\frac{1}{2}=0\end{cases}\)\(\Rightarrow\begin{cases}x=-y\\y=\frac{1}{2}\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{-1}{2}\\y=\frac{1}{2}\end{cases}\)

Vậy \(x=\frac{-1}{2};y=\frac{1}{2}\)