Ta có:\(B=5x-2\sqrt{x}+7\)
Mà \(x\ge0\)
\(\Rightarrow B=5x-2\sqrt{x}+7\ge7\)
Dấu "=" xảy ra khi x=0
Vậy MinB=7 khi x=0
ĐKXĐ:...
\(B=5\left(x-2.\frac{1}{5}\sqrt{x}+\frac{1}{25}\right)+\frac{34}{5}=5\left(\sqrt{x}-\frac{1}{5}\right)^2+\frac{34}{5}\ge\frac{34}{5}\)
\(B_{min}=\frac{34}{5}\) khi \(\sqrt{x}=\frac{1}{5}\Rightarrow x=\frac{1}{25}\)