$a)$
$C=\left (\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1} \right ).\left (\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x} \right )$ $(x \geq 0 ; x \neq 1)$
$= \left [\dfrac{2x+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\dfrac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)} \right ].\left [\dfrac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}+1}-\sqrt{x} \right ]$
$=\dfrac{2x+1-x+\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.(x-\sqrt{x}+1-\sqrt{x})$
$=\dfrac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.(x-2\sqrt{x}+1)$
$=\dfrac{1}{\sqrt{x}-1}.(\sqrt{x}-1)^2$
$=\sqrt{x}-1$
Vậy $C=\sqrt{x}-1$ với $(x \geq 0 ; x \neq 1)$
$b)$ $x=8-2\sqrt{7}$ ( Thỏa mãn ĐKXĐ )
$⇒C=\sqrt{8-2\sqrt{7}}-1$
$=\sqrt{(\sqrt{7}-1)^2}-1$
$=|\sqrt{7}-1|-1$
$=\sqrt{7}-2$
Vậy $C=\sqrt{7}-2$ với $x=8-2\sqrt{7}$
