ĐK: x\(\ge\)0
Để B đạt GTLN thì \(x-\sqrt{x}+1\) đạt GTNN
Ta có:
\(x-\sqrt{x}+1=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\\ =\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\\ \Rightarrow GTNN=\frac{3}{4}khix=\frac{1}{4}\)
Khi đó \(MaxB=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)