a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{x+1}-\dfrac{2}{y-1}=1\\\dfrac{4}{x+1}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{12}{x+1}-\dfrac{8}{y-1}=4\\\dfrac{12}{x+1}-\dfrac{9}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-1}=1\\\dfrac{4}{x+1}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=1\\\dfrac{4}{x+1}=1+\dfrac{3}{y-1}=1+3=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=1\\x+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=0\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\y\ne\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{3}{2y-1}=6\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{x-2}=7\\\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=\dfrac{5}{7}\\\dfrac{1}{2y-1}=2-\dfrac{1}{x-2}=2-1:\dfrac{5}{7}=2-\dfrac{7}{5}=\dfrac{3}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{7}+2=\dfrac{19}{7}\\2y-1=\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\\2y=\dfrac{8}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{19}{7}\\y=\dfrac{4}{3}\end{matrix}\right.\left(nhận\right)\)
c: ĐKXĐ: x<>-y và y<>1
\(\left\{{}\begin{matrix}\dfrac{4}{x+y}+\dfrac{1}{y-1}=5\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{8}{x+y}+\dfrac{2}{y-1}=10\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+y}=9\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=1\\\dfrac{2}{y-1}=\dfrac{1}{x+y}+1=\dfrac{1}{1}+1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=1\\y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1-y=1-2=-1\end{matrix}\right.\left(nhận\right)\)
d: ĐKXĐ: x<>2 và y<>1/2
\(\left\{{}\begin{matrix}\dfrac{x}{x-2}+\dfrac{3}{2y-1}=5\\\dfrac{3x}{x-2}-\dfrac{4}{2y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x-2+2}{x-2}+\dfrac{3}{2y-1}=5\\\dfrac{3x-6+6}{x-2}-\dfrac{4}{2y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{3}{2y-1}=5-1=4\\\dfrac{6}{x-2}-\dfrac{4}{2y-1}=2-3=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{8}{x-2}+\dfrac{12}{2y-1}=16\\\dfrac{18}{x-2}-\dfrac{12}{2y-1}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{26}{x-2}=13\\\dfrac{2}{x-2}+\dfrac{3}{2y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=2\\\dfrac{3}{2y-1}=4-\dfrac{2}{x-2}=4-\dfrac{2}{2}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=2\\2y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\left(nhận\right)\)
e: ĐKXĐ: x<>1 và y<>-2
\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=5-\dfrac{2x}{x-1}=5-2\cdot2=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-2=x\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\left(nhận\right)\)
