Trả lời:
\(A=\dfrac{2a+4}{a\sqrt{a}-1}+\dfrac{\sqrt{a}+2}{a+\sqrt{a}+1}-\dfrac{2}{\sqrt{a}-1}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\dfrac{2a+4}{\left(\sqrt{a}\right)^3-1^3}+\dfrac{\sqrt{a}+2}{a+\sqrt{a}+1}-\dfrac{2}{\sqrt{a}-1}\)
\(=\dfrac{2a+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{\sqrt{a}+2}{a+\sqrt{a}+1}-\dfrac{2}{\sqrt{a}-1}\)
\(=\dfrac{2a+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{2a+4+a-\sqrt{a}+2\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}=\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\)
