Question

giải hệ phương trình : \(\left\{{}\begin{matrix}\dfrac{4}{x+y}+\dfrac{1}{y-1}=5\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\dfrac{4}{x+y}+\dfrac{1}{y-1}=5\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\)

Đặt a = \(\dfrac{1}{x+y}\); b = \(\dfrac{1}{y-1}\)

=> pt <=> \(\left\{{}\begin{matrix}4a+b=5\\a-2b=-1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}b=5-4a\\a-2\left(5-4a\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}b=5-4a\\a-10+8a=-1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}b=5-4a\\9a=9\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}b=5-4.1=1\\a=1\end{matrix}\right.\)