+ Ta có:
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{(\sqrt 2)^2+ \sqrt 2}{1+ \sqrt 2}=\dfrac{\sqrt{2}(\sqrt{2}+1)}{1+\sqrt{2}}\)
\(=\dfrac{\sqrt 2(1+ \sqrt 2)}{\sqrt 2}=\sqrt{2}\).
+ Ta có:
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{3.5}-\sqrt{5.1}}{1-\sqrt{3}}\)\(=\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{5}.1}{1-\sqrt{3}}\)
\(=\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}=\dfrac{-\sqrt{5}(1-\sqrt{3})}{1-\sqrt{3}}=-\sqrt{5}\).
+ Ta có:
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{(\sqrt 2)^2.\sqrt 3-\sqrt 6}{\sqrt{4.2}- 2}\)
\(=\dfrac{\sqrt 2.(\sqrt 2.\sqrt 3)-\sqrt 6}{2\sqrt 2 -2}\)\(=\dfrac{2\sqrt{6}-\sqrt 6}{2(\sqrt{2}-1)}\)
\(=\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}=\dfrac{\sqrt{6}}{2}\).
+ Ta có:
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{(\sqrt a)^2-\sqrt a .1}{1-\sqrt a}=\dfrac{\sqrt{a}(\sqrt{a}-1)}{1-\sqrt{a}}\)
\(=\dfrac{-\sqrt{a}(1-\sqrt{a})}{1-\sqrt{a}}=-\sqrt{a}\).
+ Ta có:
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{(\sqrt p)^2-2.\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}(\sqrt{p}-2)}{\sqrt{p}-2}=\sqrt{p}\).
