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\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+........+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\frac{1}{\sqrt{1}\sqrt{2}\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{\sqrt{2}\sqrt{3}\left(\sqrt{2}+\sqrt{3}\right)}+........+\frac{1}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2009}+\sqrt{2010}\right)}\)
\(=\frac{\left(\sqrt{2010}-\sqrt{2009}\right)\left(\sqrt{2010}+\sqrt{2009}\right)}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2010}+\sqrt{2009}\right)}+.......+\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}=1-\frac{1}{\sqrt{2010}}=1-\frac{\sqrt{2010}}{2010}\)
C/m: \(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\sqrt{k+1}-\sqrt{k}\)\(\left(k\ge1,k\in\text{ℕ}\right)\)
Có: \(\dfrac{1}{\sqrt{k-1}+\sqrt{k}}>\dfrac{1}{\sqrt{k}+\sqrt{k+1}}\)
\(\Rightarrow\dfrac{2}{\sqrt{k-1}+\sqrt{k}}>\dfrac{1}{\sqrt{k}+\sqrt{k+1}}+\dfrac{1}{\sqrt{k-1}+\sqrt{k}}\)\(=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}=\sqrt{k+1}-\sqrt{k-1}\)
\(\Rightarrow2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\right)>\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{81}=9-1=8\)
\(\Rightarrow\dfrac{1}{\sqrt{1}+\sqrt{2}}+...+\dfrac{1}{\sqrt{79}+\sqrt{80}}>4\)(đpcm).
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
Xét:
\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\)
\(\Rightarrow B=\sqrt{81}-\sqrt{1}=8\)
Mặt khác, do \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}< \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{2}{\sqrt{1}+\sqrt{2}}\)
Tương tự: \(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}< \frac{2}{\sqrt{3}+\sqrt{4}}\) ....
\(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}< \frac{2}{\sqrt{79}+\sqrt{80}}\)
Cộng vế với vế ta được: \(2A>B=8\Rightarrow A>4\)
a: \(VT=\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{-\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}=\dfrac{-3\sqrt{6}+4\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
b: \(VT=\dfrac{\left(\sqrt{6x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right)}{\sqrt{6x}}\)
\(=1+\dfrac{1}{3}+1=2\dfrac{1}{3}\)
\(=\left[\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)\right]+\left[\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\right]\)
\(=\left(\dfrac{2+\sqrt{3}}{2}:\dfrac{2+\sqrt{3}+1}{2}\right)+\left(\dfrac{2-\sqrt{3}}{2}:\dfrac{2-\sqrt{3}+1}{2}\right)\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=1\)