ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x-2}{x+1}+\dfrac{2y+1}{y+1}=1\\\dfrac{x-1}{x+1}+\dfrac{y-2}{y+1}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x+2-4}{x+1}+\dfrac{2y+2-1}{y+1}=1\\\dfrac{x+1-2}{x+1}+\dfrac{y+1-3}{y+1}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-\dfrac{4}{x+1}+2-\dfrac{1}{y+1}=1\\1-\dfrac{2}{x+1}+1-\dfrac{3}{y+1}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-4}{x+1}-\dfrac{1}{y+1}=1-4=-3\\\dfrac{2}{x+1}+\dfrac{3}{y+1}=2-6=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-4}{x+1}-\dfrac{1}{y+1}=-3\\\dfrac{4}{x+1}+\dfrac{6}{y+1}=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y+1}=-11\\\dfrac{2}{x+1}+\dfrac{3}{y+1}=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y+1=\dfrac{-5}{11}\\\dfrac{2}{x+1}=-4-3:\dfrac{-5}{11}=-4+3\cdot\dfrac{11}{5}=\dfrac{33}{5}-4=\dfrac{13}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{16}{11}\\x+1=\dfrac{10}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{16}{11}\\x=-\dfrac{3}{13}\end{matrix}\right.\left(nhận\right)\)
ĐKXĐ: \(x\ne-1;y\ne-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-2}{x+1}+\dfrac{2y+1}{y+1}=1\\\dfrac{2x-2}{x+1}+\dfrac{2y-4}{y+1}=12\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow\dfrac{2y-4}{y+1}-\dfrac{2y+1}{y+1}=12-1\)
\(\Leftrightarrow\dfrac{-5}{y+1}=11\)
\(\Rightarrow y+1=-\dfrac{5}{11}\Rightarrow y=-\dfrac{16}{11}\)
Thế vào \(\dfrac{x-1}{x+1}+\dfrac{y-2}{y+1}=6\Rightarrow\dfrac{x-1}{x+1}+\dfrac{38}{5}=6\)
\(\Rightarrow\dfrac{x-1}{x+1}=-\dfrac{8}{5}\Rightarrow5x-5=-8x-7\)
\(\Rightarrow x=-\dfrac{2}{13}\)
