e: ĐKXĐ: \(x\ne y-2;x\ne-y+1\)
\(\left\{{}\begin{matrix}\dfrac{1}{x-y+2}+\dfrac{1}{x+y-1}=8\\\dfrac{2}{x-y+2}-\dfrac{1}{x+y-1}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x-y+2}+\dfrac{1}{x+y-1}+\dfrac{2}{x-y+2}-\dfrac{1}{x+y-1}=8+6\\\dfrac{1}{x-y+2}+\dfrac{1}{x+y-1}=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-y+2}=14\\\dfrac{1}{x+y-1}=8-\dfrac{1}{x-y+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y+2=\dfrac{3}{14}\\\dfrac{1}{x+y-1}=8-1:\dfrac{3}{14}=8-\dfrac{14}{3}=\dfrac{10}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=\dfrac{3}{14}-2=\dfrac{3}{14}-\dfrac{28}{14}=-\dfrac{25}{14}\\x+y-1=\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-\dfrac{25}{14}\\x+y=\dfrac{13}{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y+x+y=-\dfrac{25}{14}+\dfrac{13}{10}\\x+y=\dfrac{13}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-\dfrac{125}{70}+\dfrac{91}{70}=\dfrac{-34}{70}=-\dfrac{17}{35}\\x+y=\dfrac{13}{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{17}{70}\\y=\dfrac{13}{10}-x=\dfrac{13}{10}+\dfrac{17}{70}=\dfrac{91+17}{70}=\dfrac{108}{70}=\dfrac{54}{35}\end{matrix}\right.\)(nhận)
d: ĐKXĐ: \(x\ne-1;y\ne-1\)
\(\left\{{}\begin{matrix}\dfrac{2}{x+1}+\dfrac{1}{y+1}=2\\\dfrac{6}{x+1}-\dfrac{2}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x+1}+\dfrac{2}{y+1}=4\\\dfrac{6}{x+1}-\dfrac{2}{y+1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+1}+\dfrac{2}{y+1}+\dfrac{6}{x+1}-\dfrac{2}{y+1}=4+1\\\dfrac{2}{x+1}+\dfrac{1}{y+1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{10}{x+1}=5\\\dfrac{1}{y+1}=\dfrac{2}{x+1}-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=2\\\dfrac{1}{y+1}=\dfrac{2}{2}-2=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\left(nhận\right)\)
c: ĐKXĐ: \(x\ne1;y\ne-3\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+3}=3\\\dfrac{4x}{x-1}+\dfrac{1}{y+3}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+3}=3\\\dfrac{8x}{x-1}+\dfrac{2}{y+3}=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+3}+\dfrac{8x}{x-1}+\dfrac{2}{y+3}=3+10\\\dfrac{3x}{x-1}-\dfrac{2}{y+3}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{11x}{x-1}=13\\\dfrac{2}{y+3}=\dfrac{3x}{x-1}-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x-1}=\dfrac{13}{11}\\\dfrac{2}{y+3}=3\cdot\dfrac{13}{11}-3=\dfrac{39}{11}-\dfrac{33}{11}=\dfrac{6}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11x=13x-13\\y+3=\dfrac{22}{6}=\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\left(nhận\right)\)
a: ĐKXĐ: \(x\ne0;y\ne0\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{2}{y}=-1\\\dfrac{2}{x}+\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{4}{y}=-2\\\dfrac{2}{x}+\dfrac{1}{y}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{4}{y}-\dfrac{2}{x}-\dfrac{1}{y}=-2-3\\\dfrac{1}{x}-\dfrac{2}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{5}{y}=-5\\\dfrac{1}{x}=\dfrac{2}{y}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=\dfrac{2}{1}-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: \(x\ne y;x\ne-\dfrac{y}{2}\)
\(\left\{{}\begin{matrix}\dfrac{1}{x-y}+\dfrac{1}{2x+y}=2\\\dfrac{3}{x-y}-\dfrac{2}{2x+y}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-y}+\dfrac{2}{2x+y}=4\\\dfrac{3}{x-y}-\dfrac{2}{2x+y}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-y}+\dfrac{2}{2x+y}+\dfrac{3}{x-y}-\dfrac{2}{2x+y}=4-2\\\dfrac{1}{x-y}+\dfrac{1}{2x+y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{x-y}=2\\\dfrac{1}{2x+y}=2-\dfrac{1}{x-y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{5}{2}\\\dfrac{1}{2x+y}=2-1:\dfrac{5}{2}=2-\dfrac{2}{5}=\dfrac{8}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=\dfrac{5}{2}\\2x+y=\dfrac{5}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y+2x+y=\dfrac{5}{2}+\dfrac{5}{8}\\x-y=\dfrac{5}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=\dfrac{25}{8}\\y=x-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{24}\\y=\dfrac{25}{24}-\dfrac{5}{2}=\dfrac{25}{24}-\dfrac{60}{24}=-\dfrac{35}{24}\end{matrix}\right.\left(nhận\right)\)
