a: ĐKXĐ: \(x\ne0;y\ne0\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)
Hệ phương trình sẽ trở thành: \(\left\{{}\begin{matrix}a+b=\dfrac{3}{4}\\5a+6b=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5a+5b=\dfrac{15}{4}\\5a+6b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a+6b-5a-5b=4-\dfrac{15}{4}\\a+b=\dfrac{3}{4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{4}\\\dfrac{1}{y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: \(x\ne0;y\ne0\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}2a-5b=1\\-a+3b=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-5b=1\\-2a+6b=-\dfrac{4}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2a-5b-2a+6b=1-\dfrac{4}{5}\\2a-5b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{5}\\2a=5b+1=5\cdot\dfrac{1}{5}+1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=1\\b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=1\\\dfrac{1}{y}=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\left(nhận\right)\)
c: ĐKXĐ: \(x\ne2;y\ne1\)
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{y-1}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}a+b=-1\\2a-3b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=-2\\2a-3b=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3a+3b+2a-3b=-2+8\\a+b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=6\\b=-1-a\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\dfrac{6}{5}\\b=-1-\dfrac{6}{5}=-\dfrac{11}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{6}{5}\\\dfrac{1}{y-1}=-\dfrac{11}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=\dfrac{5}{6}\\y-1=-\dfrac{5}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{17}{6}\\y=-\dfrac{5}{11}+1=\dfrac{6}{11}\end{matrix}\right.\left(nhận\right)\)
d:ĐKXĐ: \(y\ne2x;y\ne\dfrac{x}{2}\)
Đặt \(\dfrac{1}{2x-y}=a;\dfrac{1}{x-2y}=b\)
Hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}2a+3b=\dfrac{1}{2}\\2a-b=\dfrac{1}{18}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+3b-2a+b=\dfrac{1}{2}-\dfrac{1}{18}\\2a-b=\dfrac{1}{18}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4b=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\2a=b+\dfrac{1}{18}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{9}\\a=\dfrac{b}{2}+\dfrac{1}{36}=\dfrac{1}{9}+\dfrac{1}{36}=\dfrac{5}{36}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y=\dfrac{36}{5}\\x-2y=\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y=14,4\\x-2y=4,5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-2y-x+2y=14,4-4,5\\2x-y=7,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=9,9\\y=2x-7,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,3\\y=2\cdot3,3-7,2=-0,6\end{matrix}\right.\left(nhận\right)\)