\(A=\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\left(2\sqrt{5}+2-\sqrt{2}\right)=\dfrac{8\left(\sqrt{5}+1\right)}{5-1}-2\sqrt{5}-2+\sqrt{2}\)
\(=2\left(\sqrt{5}+1\right)-2\sqrt{5}-2+\sqrt{2}=2\sqrt{5}+2-2\sqrt{5}-2+\sqrt{2}\)
\(=\sqrt{2}\)
\(B=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b. Ta có:
\(B=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}\)
Do \(\dfrac{1}{\sqrt{a}}>0;\forall a>0;a\ne1\Rightarrow1-\dfrac{1}{\sqrt{a}}< 1\)
Hay \(B< 1\)
Mà \(A=\sqrt{2}>1\)
\(\Rightarrow A>B\)
