\(Đk:x\ge0\)
a.\(A=\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}=\dfrac{\dfrac{1}{2}\left(2\sqrt{x}+1\right)+\dfrac{1}{2}}{2\sqrt{x}+1}=\dfrac{1}{2}+\dfrac{1}{2\left(2\sqrt{x}+1\right)}\)
- Do \(2\sqrt{x}+1\ge2.0+1=1\Rightarrow\dfrac{1}{2\left(2\sqrt{x}+1\right)}\le\dfrac{1}{2.1}=\dfrac{1}{2}\)
\(\Rightarrow A\le\dfrac{1}{2}+\dfrac{1}{2}=1\)
- Vậy \(MaxA=1\), đạt tại \(x=0\)
b. \(B=\dfrac{\sqrt{x}}{x-2\sqrt{x}+9}\)
- Với \(x=0\Rightarrow B=0\)
- Với \(x\ne0\Rightarrow B=\dfrac{1}{\sqrt{x}-2+\dfrac{9}{\sqrt{x}}}=\dfrac{1}{\left(\sqrt{x}+\dfrac{9}{\sqrt{x}}\right)-2}\le\dfrac{1}{2.\sqrt{\sqrt{x}.\dfrac{9}{\sqrt{x}}}-2}=\dfrac{1}{4}\)
- Vậy \(MinB=\dfrac{1}{4}\), đạt tại: \(\sqrt{x}=\dfrac{9}{\sqrt{x}}\Leftrightarrow x=9\)
c. \(C=\dfrac{\sqrt{x}-1}{x-3\sqrt{x}+11}=\dfrac{-\dfrac{1}{5}\left(x-8\sqrt{x}+16\right)+\dfrac{1}{5}\left(x-3\sqrt{x}+11\right)}{x-3\sqrt{x}+11}\)
\(=\dfrac{1}{5}-\dfrac{1}{5}.\dfrac{\left(\sqrt{x}-4\right)^2}{x-3\sqrt{x}+11}\le\dfrac{1}{5}\)
- Vậy \(MinC=\dfrac{1}{5}\), đạt tại \(\sqrt{x}-4=0\Leftrightarrow x=16\)
