\(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\ge0\\\left(3y+10\right)^{2012}\ge0\end{matrix}\right.\Leftrightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)
Mà \(\left|2x-27\right|^{2017}+\left(3y+10\right)^{2012}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-27\right|^{2011}=0\\\left(3y+10\right)^{2012}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=13,5\\y=\dfrac{-10}{3}\end{matrix}\right.\)
Vậy...
\(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\)
\(\left|2x-27\right|^{2011}\ge0;\left(3y+10\right)^{2012}\ge0\)
Dấu "=" xảy ra khi:
\(\left|2x-27\right|^{2011}=0\)
\(\Rightarrow\left|2x-27\right|=0\Rightarrow2x-27=0\Rightarrow2x=27\Rightarrow x=\dfrac{27}{2}\)
\(\left(3y+10\right)^{2012}=0\)
\(\Rightarrow3y+10=0\Rightarrow3y=-10\Rightarrow y=\dfrac{-10}{3}\)
Ta thấy:
\(\left|2x-27\right|\ge0\Rightarrow\left|2x-27\right|^{2011}\ge0\\ \left(3y+10\right)^{2012}\text{ có mũ chẵn}\Rightarrow\left(3y+10\right)^{2012}\ge0\\ \Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)
Dấu \("="\) xảy ra khi
\(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}=0\\\left(3y+10\right)^{2012}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\left|2x-27\right|=0\\3y+10=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x=27\\3y=10\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=\dfrac{10}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{27}{2}\), \(y=\dfrac{10}{3}\)
Vì \(\left|2x-27\right|^{2011}\ge0;\left(3y+10\right)^{2012}\ge0\) mà
\(\left|2x-27\right|^{2017}+\left(3y+10\right)^{2012}=0\) nên ta có:
\(\Rightarrow\left[{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=27\\3y=-10\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{27}{2}\\y=\dfrac{-10}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{27}{2};y=\dfrac{-10}{3}\)
{|2x−27|2011≥0(3y+10)2012≥0⇔|2x−27|2011+(3y+10)2012≥0{|2x−27|2011≥0(3y+10)2012≥0⇔|2x−27|2011+(3y+10)2012≥0
Mà |2x−27|2017+(3y+10)2012=0|2x−27|2017+(3y+10)2012=0
⇒{|2x−27|2011=0(3y+10)2012=0⇒⎧⎩⎨x=13,5y=−103
