Question

Answer 1

\(\left\{{}\begin{matrix}2\sqrt{x-1}+\dfrac{1}{y-3}=5.\\5\sqrt{x-1}+\dfrac{3}{y-3}=13.\end{matrix}\right.\) \(\left(x\ge1;y\ne3\right).\)

Đặt \(\sqrt{x-1}=a;\dfrac{1}{y-3}=b\left(b\ne0\right).\)

\(\Rightarrow\left\{{}\begin{matrix}2a+b=5.\\5a+3b=13.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2.\\b=1.\end{matrix}\right.\) (TM).

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=2.\\\dfrac{1}{y-3}=1.\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x=5.\\y=4.\end{matrix}\right.\) (TM).

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(5;4\right).\)