\(a.ĐK:x\ge0;x\ne1\)
\(M=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}+\dfrac{1}{x+1}\right].\dfrac{x+1}{\sqrt{x}-1}\)
\(=\left[\dfrac{\sqrt{x}}{x+1}+\dfrac{1}{x+1}\right].\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{x+1}.\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b)\(M\sqrt{x}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+2}{\sqrt{x}-1}\)
\(=\sqrt{x}+2+\dfrac{2}{\sqrt{x}-1}=\left(\sqrt{x}-1+\dfrac{2}{\sqrt{x}-1}\right)+3\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{2}{\sqrt{x}-1}}+3=2\sqrt{2}+3\)
\(\Rightarrow M\sqrt{x}\ge2\sqrt{2}+3\)
Dấu "=" xảy ra khi \(\sqrt{x}-1=\dfrac{2}{\sqrt{x}-1}\Leftrightarrow\sqrt{x}=\sqrt{2}+1\) (tm)
Vậy GTNN của \(M\sqrt{x}=2\sqrt{2}+3\)
