ĐKXĐ: \(x\ne4;y\ne-1\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-4}=u\\\dfrac{1}{y+1}=u\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}3u+2v=\dfrac{15}{12}\\2u-v=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+2v=\dfrac{15}{12}\\4u-2v=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7u=-\dfrac{11}{4}\\v=2u+2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=-\dfrac{11}{28}\\v=\dfrac{17}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-4}=-\dfrac{11}{28}\\\dfrac{1}{y+1}=\dfrac{17}{14}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-4=-\dfrac{28}{11}\\y+1=\dfrac{14}{17}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{16}{11}\\y=-\dfrac{3}{17}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{x-4}+\dfrac{2}{y+1}=\dfrac{15}{12}\\\dfrac{2}{x-4}-\dfrac{1}{y+1}=-2\end{matrix}\right.\)
Đặt: \(\left\{{}\begin{matrix}a=\dfrac{1}{x-4}\\b=\dfrac{1}{y+1}\end{matrix}\right.\)
Hệ phương trình: \(\Leftrightarrow\left\{{}\begin{matrix}3a+2b=\dfrac{15}{12}\\2a-b=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a+2b=\dfrac{15}{12}\\4a-2b=-4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7a=-\dfrac{11}{4}\\2a-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\2\cdot\left(-\dfrac{11}{28}\right)-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\-\dfrac{11}{14}-b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{11}{28}\\b=\dfrac{17}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-4}=-\dfrac{11}{28}\\\dfrac{1}{y+1}=\dfrac{17}{14}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-4=\dfrac{1}{-\dfrac{11}{28}}\\y+1=\dfrac{1}{\dfrac{17}{14}}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{16}{11}\\y=-\dfrac{3}{17}\end{matrix}\right..}\)
