Ta có: \(\left(x-y+1\right)^{2010}\ge0\forall x,y\)
\(\left|2y-1\right|\ge0y\forall\)
\(\Rightarrow\left(x-y +1\right)^{20010}+\left|2y-1\right|\ge0\forall x,y\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}x-y+1=0\\\left|2y-1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=-1\\2y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1+y\\y=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-3}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\).
Từ (x-y+1)2010+\(\left|2y-1\right|\)=0.
Suy ra \(\left[{}\begin{matrix}\left(x-y+1\right)^{2010}=0\\\left|2y-1\right|=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x-y+1=0\\2y-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x-y=-1\\y=\left(0+1\right):2=\dfrac{1}{2}\end{matrix}\right.\)
y=\(\dfrac{1}{2}\)\(\Rightarrow\)x=y+(-1)=\(\dfrac{1}{2}\)-1=\(\)\(\dfrac{-1}{2}\)
