Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\).Ta có:
\(B\ge\sqrt{5x-4+12-5x}=\sqrt{-\left(4-12\right)}=\sqrt{8}=\sqrt{4}.\sqrt{2}=2\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{5x-4}\ge0\\\sqrt{12-5x}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}5x\ge4\\5x\le12\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{4}{5}\\x\le\frac{12}{5}\end{cases}\Leftrightarrow\frac{4}{5}\le x\le\frac{12}{5}}\)
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