Ta có: \(\dfrac{1}{\sqrt{a}+\sqrt{a+1}}=\dfrac{\sqrt{a}-\sqrt{a+1}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a}-\sqrt{a+1}\right)}=\dfrac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}\)
Áp dụng:
\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\\ =\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+...+\sqrt{36}-\sqrt{35}\\ =\sqrt{36}-\sqrt{4}=6-2=4\)

