Question

\(\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{3}{y+2}=\dfrac{1}{2}\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=\dfrac{1}{18}\end{matrix}\right.\) Làm chi tiết giúp ạ! 

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{y+2}+\dfrac{1}{y+2}=\dfrac{1}{2}-\dfrac{1}{18}\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=\dfrac{1}{18}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\\dfrac{2}{x-1}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\x-1=12\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(13;7\right)\)