Question

[Hoàng Minh Tiến]

\(\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-2005\right)^2}< hoac=0\)

Answer 1

[Đinh Đức Hùng]

Vì \(\sqrt{\left(x-y\right)^2}=\left|x-y\right|\ge0\forall x;y\)

\(\sqrt{\left(y-2015\right)^2}=\left|y-2016\right|\ge0\forall y\)

\(\Rightarrow\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-2015\right)^2}=\left|x-y\right|+\left|y-2015\right|\ge0\forall x;y\)

Để \(\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-2005\right)^2}\le0\Leftrightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|y-2005\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-2005=0\end{cases}\Rightarrow x=y=2005}\)

Vậy \(x=y=2005\)