Rút gọn biều

Ta có:

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

Khi đó $\frac{1}{\sqrt{2}-\sqrt{3}}=-(\sqrt{2}+\sqrt{3}) ; \frac{1}{\sqrt{3}-\sqrt{4}}=-(\sqrt{3}+\sqrt{4})$

$\frac{1}{\sqrt{2 n}-\sqrt{2 n+1}}=-(\sqrt{2 n}+\sqrt{2 n+1})$

$\Rightarrow P=\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}-\ldots+\frac{1}{\sqrt{2 n}-\sqrt{2 n+1}}$

$=-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-(\sqrt{4}+\sqrt{5})+\ldots .-(\sqrt{2 n}+\sqrt{2 n+1})$

$=-(\sqrt{2}+\sqrt{2 n+1})$