a) Ta có: \(\left|x\right|=\left|y\right|\)
\(\Rightarrow\left|x\right|=\left[{}\begin{matrix}-x\\x\end{matrix}\right.\)
\(\Rightarrow\left|y\right|=\left[{}\begin{matrix}y\\-y\end{matrix}\right.\)
Mà \(x< 0;y>0\)
\(\Rightarrow x+y=0\)
Ta có: \(\left|x\right|=\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{x}\right|=\left[{}\begin{matrix}\dfrac{1}{x}\\-\dfrac{1}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{y}=\left[{}\begin{matrix}\dfrac{1}{y}\\-\dfrac{1}{y}\end{matrix}\right.\)
Mà \(x< 0;y>0\)
\(\dfrac{1}{x}+\dfrac{1}{y}=0\)