Simplifying Radical expressions

Question

Simplify $\dfrac{1}{\sqrt 1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\dfrac{1}{\sqrt{8}+\sqrt{9}}$

Answer 1

Hint:

$$\frac1{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}=\sqrt{n+1}-\sqrt{n}$$

Then $$\frac{1}{\sqrt 1+\sqrt{2}}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+\cdots +\frac{1}{\sqrt8+\sqrt9}=$$ $$=\sqrt2-\sqrt1+\sqrt3-\sqrt2+\sqrt4-\sqrt3+...+\sqrt9-\sqrt8=\sqrt9-\sqrt1=3-1=2$$

Answer 2

Hint:)

Use $$\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=-\sqrt{n}+\sqrt{n+1}$$ and make a telescopic expression.