Vì \(|3x^2-27|\ge0\)\(\forall x\)\(\Rightarrow|3x^2-27|^{2019}\ge0\)\(\forall x\)
\(\left(5y+12\right)^{2018}\ge0\)\(\forall y\)
\(\Rightarrow|3x^2-27|^{2019}+\left(5y+12\right)^{2018}\ge0\)\(\forall x,y\)
mà \(|3x^2-27|^{2019}+\left(5y+12\right)^{2018}=0\)
\(\Rightarrow\)Dấu = chỉ xảy ra khi \(|3x^2-27|^{2019}=0\)và \(\left(5y+12\right)^{2018}=0\)
\(\Rightarrow|3x^2-27|=0\)và \(5y+12=0\)
\(\Rightarrow3x^2-27=0\)và \(5y=-12\)
\(\Rightarrow3x^2=27\)và \(y=\frac{-12}{5}\)
\(\Rightarrow x^2=9\)và \(y=\frac{-12}{5}\)
\(\Rightarrow x=3\)hoặc \(x=-3\)và \(y=\frac{-12}{5}\)