b.
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{2x-1}\ge0\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x-1}+x\ge0+\dfrac{1}{2}=\dfrac{1}{2}\)
\(y_{min}=\dfrac{1}{2}\) khi \(x=\dfrac{1}{2}\)
b.
ĐKXĐ: \(x>0\)
\(y=\dfrac{x}{\sqrt{x}}+\dfrac{2011\sqrt{x}}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+2011\ge2\sqrt{\dfrac{\sqrt{x}}{\sqrt{x}}}+2011=2013\)
\(y_{min}=2013\) khi \(\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\)
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