2) \(\dfrac{\left(1+\sqrt{a}\right)^2-\left(2-\sqrt{a}\right)^2}{1-2\sqrt{a}}:\dfrac{\sqrt{a}}{3}\left(a>0,a\ne\dfrac{1}{4}\right)\)
\(=\dfrac{\left(1+\sqrt{a}-2+\sqrt{a}\right)\left(1+\sqrt{a}+2-\sqrt{a}\right)}{1-2\sqrt{a}}.\dfrac{3}{\sqrt{a}}\)
\(=\dfrac{3.\left(2\sqrt{a}-1\right)}{1-2\sqrt{a}}.\dfrac{3}{\sqrt{a}}=-\dfrac{9}{\sqrt{a}}\)
5) \(\left(5-\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}\right)\left(2-\dfrac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)\left(a\ge0\right)\)
\(=\left(5-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\right)\left(2-\dfrac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)
\(=\left(5-\sqrt{a}\right)\left(2-\sqrt{a}\right)=10-7\sqrt{a}+a\)
6) \(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)\left(a,b\ge0,a\ne9,b\ne25\right)\)
\(=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2+\dfrac{\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)
3) Ta có: \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\sqrt{a}-2\)
=0
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