Question

giải hpt \(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-3\right)^2=1\\\left(x-1\right)\left(y-3\right)+3=x+y\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-3\right)^2=1\\\left(x-1\right)\left(y-3\right)-\left(x-1\right)-\left(y-3\right)+1=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x-1=a\\y-3=b\end{matrix}\right.\)

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

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