Question

\(\left\{{}\begin{matrix}|x-1|+\dfrac{2}{y}=18\\-|x-1|+\dfrac{4}{y}=1\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(y\ne0\)

Đặt \(\left\{{}\begin{matrix}\left|x-1\right|=u\ge0\\\dfrac{1}{y}=v\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}u+2v=18\\-u+4v=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{35}{3}\\v=\dfrac{19}{6}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=\dfrac{35}{3}\\\dfrac{1}{y}=\dfrac{19}{6}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{38}{3}\\x=-\dfrac{32}{3}\end{matrix}\right.\\y=\dfrac{6}{19}\end{matrix}\right.\)