Lời giải:
a) ĐKXĐ
\(\left\{\begin{matrix} x\geq 0\\ x-4\neq 0\\ x+2\sqrt{x}\neq 0\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 4\\ \sqrt{x}(\sqrt{x}+2)\neq 0\end{matrix}\right.\Rightarrow x>0; x\neq 4\)
\(C=\left(\frac{\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}+\frac{1}{(\sqrt{x}-2)(\sqrt{x}+2)}\right):\frac{1}{\sqrt{x}(\sqrt{x}+2)}\)
\(=\frac{\sqrt{x}-1}{(\sqrt{x}-2)(\sqrt{x}+2)}.\sqrt{x}(\sqrt{x}+2)\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-2}\)
b) Khi $x=\frac{1}{9}$ thì \(\sqrt{x}=\frac{1}{3}\)
\(C=\frac{\frac{1}{3}(\frac{1}{3}-1)}{\frac{1}{3}-2}=\frac{2}{15}\)
c) Vì $x>4$ nên \(\sqrt{x}-2>0\)
\(C=\frac{x-\sqrt{x}}{\sqrt{x}-2}=\frac{x-2\sqrt{x}+\sqrt{x}-2+2}{\sqrt{x}-2}=\sqrt{x}+1+\frac{2}{\sqrt{x}-2}\)
\(=(\sqrt{x}-2)+\frac{2}{\sqrt{x}-2}+3\)
\(\geq 2\sqrt{(\sqrt{x}-2).\frac{2}{\sqrt{x}-2}}+3=2\sqrt{2}+3\) (áp dụng BĐT Cô-si cho 2 số dương )
Vậy \(C_{\min}=2\sqrt{2}+3\)
