Vì \(\left(x-5\right)^{2018}\ge0;\left|2y^2-162\right|^{2018}\ge0\Rightarrow\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}\ge0\)
mà \(\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}=0\)
Dấu ''='' xảy ra khi x = 5 ; \(2y^2=162\Leftrightarrow y^2=81\Leftrightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Vì \(\left(x-5\right)^{2018}\ge0\\ \left|2y^2-162\right|^{2018}\ge0\\ \)
Suy ra phương trình dc thỏa mãn khi và chỉ khi x-5 = 0 và 2y^2-162=0
\(\left\{{}\begin{matrix}\left(x-5\right)^{2018}=0\\\left|2y^2-162\right|^{2018}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\2\left(y^2-81\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\pm9\end{matrix}\right.\)