ĐKXĐ: x≠0,x≠1,x>0
\(A=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\dfrac{a-1}{\sqrt{a}}\right)\left(\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\left(\dfrac{\left(a-1\right)\left(a+2\sqrt{a}+1+a-2\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)}\right)=\dfrac{2\sqrt{a}}{\sqrt{a}}+\dfrac{2a+2}{\sqrt{a}}=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
