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\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\sqrt{2}-1\)
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: \(2A>B=8\Rightarrow A>4\)
Ta có :Đặt t = \(\sqrt{3+\sqrt{3+\sqrt{3+...+\sqrt{3}}}} ( 2014 dấu căn )\)
\(\Rightarrow\) t > \(\sqrt{3} > \sqrt{1} = 1\)
\(\Rightarrow\) \(\sqrt{3+\sqrt{3+\sqrt{3+...+\sqrt{3}}}}\)(2013 dấu căn ) = \(t^2 -3\)
Do đó : \(M = \frac{3-t}{6-( t^2 - 3 )}\)= \(\frac{3-t}{9-t^2}\) = \(\frac{3-t}{(3-t)(3+t)}\) = \(\frac{1}{3+t}\)
Vì t>1 \(\Rightarrow\) 3+t > 4 \(\Rightarrow\) \(\frac{1}{3+t}\) < \(\frac{1}{4}\)
Vậy M < \(\frac{1}{4}\)
