Ta thấy: \(\left\{{}\begin{matrix}\left(3x+1\right)^6\ge0\\\left|1-y\right|\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2\left(3x+1\right)^6\ge0\\3\left|1-y\right|^3\ge0\end{matrix}\right.\)
\(\Rightarrow2\left(3x+1\right)^6+3\left|1-y\right|^3\ge0\)
\(\Rightarrow2\left(3x+1\right)^6+3\left|1-y\right|^3+2\ge2\)
\(\Rightarrow\dfrac{1}{2\left(3x+1\right)^6+3\left|1-y\right|^3+2}\le\dfrac{1}{2}\)
\(\Rightarrow\dfrac{3}{2\left(3x+1\right)^6+3\left|1-y\right|^3+2}\le\dfrac{3}{2}\)
Xảy ra khi \(\left\{{}\begin{matrix}2\left(3x+1\right)^6=0\\3\left|1-y\right|^3=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=1\end{matrix}\right.\)
Đặt:
\(A=\dfrac{3}{2\left(3x+1\right)^6+3\left|1-y\right|^3+2}\)
\(\left\{{}\begin{matrix}\left(3x+1\right)^6\ge0\Rightarrow2\left(3x+1\right)^6\ge0\forall x\\\left|1-y\right|\ge0\Rightarrow\left|1-y\right|^3\ge0\Rightarrow3\left|1-y\right|^3\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow2\left(3x+1\right)^6+3\left|1-y\right|^3\ge0\)
\(\Rightarrow2\left(3x+1\right)^6+3\left|1-y\right|^3+2\ge2\)
\(\Rightarrow A\le\dfrac{3}{2}\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2\left(3x+1\right)^6=0\Rightarrow3x+1=0\Rightarrow3x=-1\Rightarrow x=\dfrac{-1}{3}\\3\left|1-y\right|^3=0\Rightarrow1-y=0\Rightarrow y=1\end{matrix}\right.\)
Vậy \(MAX_A=\dfrac{3}{2}\) khi \(x=\dfrac{-1}{3};y=1\)
