Question

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

Answer 1

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

Do \(\left(x-1\right)^2+y^2=1\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|\le1\\\left|y\right|\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^3\le\left(x-1\right)^2\\y^3\le y^2\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^3+y^3\le1\)

Dấu "=" xảy ra khi và chỉ khi: \(\left[{}\begin{matrix}x-1=1\\y=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)=\left(2;0\right);\left(1;1\right)\)