Question

Rút gọn

\(A=\frac{1}{1+\sqrt{2}}\)+\(\frac{1}{\sqrt{2}+\sqrt{3}}\)+\(\frac{1}{\sqrt{3}+\sqrt{4}}\)+....+\(\frac{1}{\sqrt{2014}+\sqrt{2015}}\)

Answer 1

\(\frac{1}{1+\text{ }\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2014}+\sqrt{2015}}\)

\(=\frac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+..+\frac{\sqrt{2014}-\sqrt{2015}}{\left(\sqrt{2014}+\sqrt{2015}\right)\left(\sqrt{2014}-\sqrt{2015}\right)}\)

\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{2014}-\sqrt{2015}}{2014-2015}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2005}-\sqrt{2004}=\sqrt{2005}-1\)

Answer 2

dangj tổng quát : cmr :\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n}-\sqrt{n+1}\left(\right)dùngtrụccăthứcởmẫu\left(\right)\)