a: Đặt 1/x=a
1/(y-2)=b
Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}2a+3b=4\\4a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+6b=8\\4a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5b=7\\4a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{-1}{10}\\\dfrac{1}{y-2}=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y-2=\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=\dfrac{19}{7}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{y}=0\\\dfrac{2}{x+1}+\dfrac{3}{y}=-1\end{matrix}\right.\)
=>Hệ phương trình vô nghiệm
a) \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4.\\\dfrac{4}{x}+\dfrac{1}{y-2}=1.\end{matrix}\right.\) \(ĐK:x\ne0;y\ne2.\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y-2}=b\left(a;b\ne0\right).\)
\(\Rightarrow\left\{{}\begin{matrix}2a+3b=4.\\4a+b=1.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{-1}{10}.\\b=\dfrac{7}{5}.\end{matrix}\right.\) (TM).
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{-1}{10}.\\\dfrac{1}{y-2}=\dfrac{7}{5}.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-10\left(TM\right).\\y=\dfrac{19}{7}\left(TM\right).\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-10;\dfrac{19}{7}\right).\)
