a: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{2}{x}+\dfrac{3}{y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{2}{x}+\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}+\dfrac{2}{x}+\dfrac{3}{y}=3+5\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{x}=8\\\dfrac{1}{y}=\dfrac{1}{x}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\\dfrac{1}{y}=1:\dfrac{5}{8}-1=\dfrac{8}{5}-1=\dfrac{3}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=\dfrac{5}{3}\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{135}{x}-\dfrac{63}{y}=81\\\dfrac{28}{x}+\dfrac{63}{y}=245\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{135}{x}-\dfrac{63}{y}+\dfrac{28}{x}+\dfrac{63}{y}=81+245\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{161}{x}=326\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{9}{y}=35-\dfrac{4}{x}=35-4:\dfrac{1}{2}=27\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\left(nhận\right)\)
c: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=5\\\dfrac{2}{x}+\dfrac{5}{y}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{5}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}-\dfrac{2}{x}-\dfrac{5}{y}=10-7\\\dfrac{1}{x}+\dfrac{1}{y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{3}{y}=3\\\dfrac{1}{x}=5-\dfrac{1}{y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\\dfrac{1}{x}=5-\dfrac{1}{-1}=5+1=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-1\end{matrix}\right.\left(nhận\right)\)
d: ĐKXĐ: \(x\ne\pm y\)
\(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{5}{8}\\\dfrac{1}{x+y}-\dfrac{1}{x-y}=-\dfrac{3}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\dfrac{1}{x-y}+\dfrac{1}{x+y}-\dfrac{1}{x-y}=\dfrac{5}{8}-\dfrac{3}{8}\\\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{5}{8}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x+y}=\dfrac{2}{8}\\\dfrac{1}{x-y}=\dfrac{5}{8}-\dfrac{1}{x+y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\dfrac{1}{x-y}=\dfrac{5}{8}-\dfrac{1}{8}=\dfrac{4}{8}=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=8\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\y=x-2=5-2=3\end{matrix}\right.\left(nhận\right)\)
