Đk: \(a\ge0\)
\(\dfrac{a-\sqrt{3a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-\sqrt{3a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a-\sqrt{3a}+3\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)
Ta có: \(a\sqrt{a}+3\sqrt{3}=\left(\sqrt{a}\right)^3+\left(\sqrt{3}\right)^3=\left(\sqrt{a}+\sqrt{3}\right)\left(a+3-\sqrt{3a}\right)\)
\(\Rightarrow\dfrac{a-\sqrt{3a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-\sqrt{3a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a+3-\sqrt{3a}\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)