Ta có
\(\left|2x-0,\left(24\right)\right|+\left|3y+0,1\left(5\right)\right|=0\)
\(\Rightarrow\left|2x-\frac{24}{99}\right|+\left|3y+0,\left(5\right)-0,4\right|=0\)
\(\Rightarrow\left|2x-\frac{8}{33}\right|+\left|3y+\frac{5}{9}-\frac{4}{5}\right|=0\)
Ta có
\(\begin{cases}\left|2x-\frac{8}{33}\right|\ge0\\\left|3y+\frac{5}{9}-\frac{2}{5}\right|\ge0\end{cases}\)
\(\Rightarrow\begin{cases}2x-\frac{8}{33}=0\\3y+\frac{5}{9}-\frac{2}{5}=0\end{cases}\)
\(\Rightarrow\begin{cases}2x=\frac{8}{33}\\3y=\frac{7}{45}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{4}{33}\\y=\frac{7}{135}\end{cases}\)
Vậy \(\left(x;y\right)=\left(\frac{4}{45};\frac{7}{135}\right)\)
