Ta có :
\(2\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\right)\)
\(>\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)
\(=\frac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)
\(=\frac{1}{2}\left(\sqrt{101}-\sqrt{1}\right)>\frac{9}{2}\)
\(\Rightarrow\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}>\frac{9}{4}\left(đpcm\right)\)
Chúc bạn học tốt !!!
Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{3}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)
\(\Rightarrow2A=\frac{\left(\sqrt{3}\right)^2-\left(\sqrt{1}\right)^2}{\sqrt{1}+\sqrt{3}}+...+\frac{\left(\sqrt{99}\right)^2-\left(\sqrt{97}\right)^2}{\sqrt{97}+\sqrt{99}}\)
\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{99}-\sqrt{97}\)
\(=\sqrt{99}-1\)
\(\Rightarrow A=\frac{\sqrt{99}-1}{2}=\frac{2\sqrt{99}-2}{4}>\frac{9}{4}\left(đpcm\right)\)
