Question

\(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(x\ne1;y\ne1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-1}=u\\\dfrac{1}{y-1}=v\end{matrix}\right.\) hệ trở thành:

\(\left\{{}\begin{matrix}5u+v=10\\u-3v=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}15u+3v=30\\u-3v=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}16u=48\\v=\dfrac{u-18}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=3\\v=-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=3\\\dfrac{1}{y-1}=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=\dfrac{1}{3}\\y-1=-\dfrac{1}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)