\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+...+\dfrac{\sqrt{100}-\sqrt{99}}{1}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{100}-\sqrt{99}=\sqrt{100}-1=10-1=9\)
Đặt A = \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
= \(\dfrac{2-1}{1+\sqrt{2}}+\dfrac{3-2}{\sqrt{2}+\sqrt{3}}+...+\dfrac{100-99}{\sqrt{99}+\sqrt{100}}\)
= \(\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{1+\sqrt{2}}+\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+...+\dfrac{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}{\sqrt{99}+\sqrt{100}}\)= \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
= \(\sqrt{100}-1=10-1=9\)
