Question

[Nguyễn Đỗ Kha]

Tìm tất cả các cặp số (x;y) thỏa mãn :\(\left(2x-y+7\right)^{2012}+|x-3|^{2013}\le0\)

Answer 1

[Akai Haruma]

Lời giải:

Ta thấy:

\((2x-y+7)^{2012}=[(2x-y+7)^{1006}]^2\geq 0, \forall x,y\in\mathbb{R}\)

\(|x-3|^{2013}\geq 0, \forall x\in\mathbb{R}\)

\(\Rightarrow (2x-y+7)^{2012}+|x-3|^{2013}\geq 0, \forall x,y\)

Do đó để thỏa mãn điều kiện đề bài thì:

\((2x-y+7)^{2012}+|x-3|^{2013}=0\)

\(\Leftrightarrow \left\{\begin{matrix} (2x-y+7)^{2012}=0\\ |x-3|^{2013}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x-y+7=0\\ x=3\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=3\\ y=13\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

\(\left\{{}\begin{matrix}\left(2x-y+7\right)^{2012}\ge0\\\left|x-3\right|^{2013}\ge0\end{matrix}\right.\) \(\Rightarrow\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}\ge0\)

Vậy \(\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}\le0\Leftrightarrow\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+7=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}6-y+7=0\\x=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=13\end{matrix}\right.\)