Question

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

Answer 1

ĐKXĐ: \(x;y\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2+2=5+ab\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\a^2+b^2-ab-3=0\end{matrix}\right.\)

\(\Rightarrow a^2+\left(3-a\right)^2-a\left(3-a\right)-3=0\)

\(\Leftrightarrow3a^2-9a+6=0\Rightarrow\left[{}\begin{matrix}a=1\Rightarrow b=2\\a=2\Rightarrow b=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=2\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x-1}=2\\\sqrt{y-1}=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(2;5\right);\left(5;2\right)\)