ĐKXĐ: x<>-2
\(\left\{{}\begin{matrix}\dfrac{3}{x+2}-2\left|2y-1\right|=4\\\dfrac{2}{x+2}+\left|2y-1\right|=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+2}-2\left|2y-1\right|=4\\\dfrac{4}{x+2}+2\left|2y-1\right|=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+2}-2\left|2y-1\right|+\dfrac{4}{x+2}+2\left|2y-1\right|=4+10\\\dfrac{2}{x+2}+\left|2y-1\right|=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{7}{x+2}=14\\\left|2y-1\right|=5-\dfrac{2}{x+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=\dfrac{1}{2}\\\left|2y-1\right|=5-2:\dfrac{1}{2}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}-2=-\dfrac{3}{2}\\2y-1\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y\in\left\{1;0\right\}\end{matrix}\right.\)
