Question

\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{5}{2y-1}=3\\\dfrac{3}{x-2}-\dfrac{1}{2y-1}=1\end{matrix}\right.\)

Answer 1

Đặt 1/x-2=a;

1/2y-1=b

\(\Leftrightarrow\left\{{}\begin{matrix}a+5b=3\\3a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=3\\15a-5b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=3a-1=\dfrac{3}{2}-1=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=2\\2y-1=2\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(4;\dfrac{3}{2}\right)\)