Question

Tính : \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{2+\sqrt{3}}+...+\dfrac{1}{20+\sqrt{21}}\)

Answer 1

\(1=2-1=\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)=>\(\dfrac{1}{1+\sqrt{2}=}=\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}-1\)

cmtt thì biểu thức thành

\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{21}-\sqrt{20}\)=\(1+\sqrt{21}\)