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\(\forall n\inℕ^∗\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) (*)
Thay n=1; n=2; n=3; .....; n=2004 Ta có:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)
\(=1-\frac{1}{\sqrt{2005}}\)
Áp dụng \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n+1}\sqrt{n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
\(\Rightarrow S=\frac{1}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{2004}}{2004}-\frac{\sqrt{2005}}{2005}\)
\(=1-\frac{\sqrt{2005}}{2005}\)
CMR \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}< 2\)
Lời giải:
Xét số hạng tổng quát \(\frac{1}{(n+1)\sqrt{n}}\):
\(\frac{1}{(n+1)\sqrt{n}}=\frac{(n+1)-n}{(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}.\sqrt{n(n+1)}}\)
\(< \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\frac{\sqrt{n+1}+\sqrt{n}}{2}.\sqrt{n(n+1)}}\)
\(\Leftrightarrow \frac{1}{(n+1)\sqrt{n}}< 2.\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Cho $n=1,2,....,2004$
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{2005\sqrt{2004}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{2005\sqrt{2004}}< 2(1-\frac{1}{\sqrt{2005}})< 2\) (đpcm)
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(P=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2005\sqrt{2004}}\)
\(\Rightarrow P< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(\Rightarrow P< 2\left(1-\frac{1}{\sqrt{2005}}\right)< 2.1=2\)
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\left(\sqrt{n}-\sqrt{n+1}\right)=\sqrt{n+1}-\sqrt{n}\)
Áp dụng ta có
A = \(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+....+\sqrt{2005}-\sqrt{2004}=\sqrt{2005}-1\)
Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Từ đó ta có
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)
\(=1-\frac{1}{\sqrt{2005}}=\frac{\sqrt{2005}-1}{\sqrt{2005}}\)