\(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
ta thấy rằng:
\(\left(x-2\right)^{2012}>=0\)
\(\left|y^2-9\right|^{2014}>=0\)
Để \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
Thì (x-2)=0 và |y2 - 9|=0
=> x=2 và y= 3
(x−2)2012+∣y2−9∣2014=0
Ta thấy:
\(\left(x-2\right)^{2012}\)≥0;\(\left|y^2-9\right|^{2014}\)≥0
\(\Leftrightarrow\)\(\left(x-2\right)^{2012}=0\) ⇒\(x-2=0\Rightarrow x=2\)
\(\Leftrightarrow\)\(\left|y^2-9\right|^{2014}=0\Rightarrow y^2-9=0\)\(\rightarrow\)\(y^2=9\)
\(\Rightarrow\)\(y=\left\{{}\begin{matrix}3\\-3\end{matrix}\right.\)
Vậy:\(\left[{}\begin{matrix}x=2\\y=3\end{matrix}\right.\) hoàc \(\left[{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
