Ta có :
\(\left\{{}\begin{matrix}\left(x-2\right)^{2012}\ge0\\\left|y^2-9\right|^{2014}\ge0\end{matrix}\right.\)
Lại có : \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\\left[{}\begin{matrix}y=3\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
Vì (x-2)2012 ≥ 0
/y2 -9/2014 ≥ 0
=> (x-2)2012 + /y2 -9/2014 = 0
=> (x-2)2012 = 0
/y2 -9/2012 = 0
=> x-2 = 0
y2 -9 = 0
=> x = 2
y2 = 9
=> x=2
y = 3 hoặc -3
Vậy x=2
y = 3 hoặc -3
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\y^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\y=\pm3\end{matrix}\right.\)
