ĐKXĐ: x>0
\(B=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}\)
\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\cdot\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}=2\forall x>0\)
=>\(B=\sqrt{x}+\frac{1}{\sqrt{x}}-1\ge2-1=1\forall x>0\)
Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=1\)
=>x=1