Lời giải:
ĐK: $x\geq 0$
Ta có:
$C=x-\sqrt{x}=(\sqrt{x})^2-2.\sqrt{x}.\frac{1}{2}+(\frac{1}{2})^2-\frac{1}{4}$
$=(\sqrt{x}-\frac{1}{2})^2-\frac{1}{4}\geq 0-\frac{1}{4}$
Hay $C\geq \frac{-1}{4}$
Vậy $C_{\min}=-\frac{1}{4}$ khi $(\sqrt{x}-\frac{1}{2})^2=0$ hay $C_{\min}=\frac{-1}{4}$ khi $x=\frac{1}{4}$