ĐKXĐ : \(y\ge\frac{4}{\sqrt{3}}\) hoặc \(y\le\frac{-4}{\sqrt{3}}\)
\(B=-\left|1-2x\right|-2\left|x-3\right|-\sqrt{3y^2-16}+2021\)
\(B=-\left(\left|1-2x\right|+\left|2x-6\right|\right)-\sqrt{3y^2-16}+2021\)
\(B\le-\left|1-2x+2x-6\right|-0+2021=2016\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(1-2x\right)\left(2x-6\right)\ge0\left(1\right)\\3y^2-16=0\left(2\right)\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}1-2x\ge0\\2x-6\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{2}\\x\ge3\end{cases}}}\) ( loại )
TH2 : \(\hept{\begin{cases}1-2x\le0\\2x-6\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le3\end{cases}\Leftrightarrow}\frac{1}{2}\le x\le3}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(y^2=\frac{16}{3}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y=\sqrt{\frac{16}{3}}\\y=-\sqrt{\frac{16}{3}}\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{4}{\sqrt{3}}\\y=\frac{-4}{\sqrt{3}}\end{cases}}}\) ( nhận )
Vậy GTNN của \(B\) là \(2016\) khi \(\frac{1}{2}\le x\le3\) và \(y=\frac{4}{\sqrt{3}}\) hoặc \(y=\frac{-4}{\sqrt{3}}\)
-,-
