\(A=\dfrac{1+\sqrt{a}+a+\sqrt{a}}{\sqrt{a}+1}.\dfrac{\sqrt{a}-1-a+\sqrt{a}}{\sqrt{a}-1}++a=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\dfrac{-\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}+a=-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)+a=-a+1+a=1\)
c: Ta có: \(C=\left(\dfrac{1}{2\sqrt{x}+2}-\dfrac{1}{2\sqrt{x}-2}+\dfrac{x+1}{x^2-1}\right)\cdot\left(1-\dfrac{1}{x}\right)\)
\(=\left(\dfrac{\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x-1}{x}\)
\(=\left(\dfrac{-1}{x-1}+\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{x}\)
=0
