\(a)C=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2}{x-1}\)
\(b)x=\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}-1}-\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}+1}\\ =\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}+1\right)}{1+\sqrt{7}-1}-\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}-1\right)}{1+\sqrt{7}-1}\\ =\sqrt{1+\sqrt{7}}+1-\sqrt{1+\sqrt{7}}+1\\ =2\)
Thay \(x=2\) vào \(C\)
\(\dfrac{2}{2-1}=\dfrac{2}{1}=2\)
