Chứng minh rằng \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{2005\sqrt{20004}}< 2.\)
Chứng minh
\(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}.\sqrt{3}}+\frac{1}{\sqrt{3}.\sqrt{4}}+...+\frac{1}{\sqrt{2004}.\sqrt{2005}}< 2\)
Chứng minh rằng: \(\frac{1}{2}\)+\(\frac{1}{3\sqrt{2}}\)+\(\frac{1}{4\sqrt{2}}\)+...+\(\frac{1}{2005\sqrt{2004}}\)< 2
\(\frac{1}{n\sqrt{n-1}}=\frac{\sqrt{n-1}}{\left(n-1\right)n}=\sqrt{n-1}.\frac{1}{\left(n-1\right)n}=\sqrt{n-1}\left(\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\sqrt{n-1}\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{n}}\right)\)
\(=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\frac{\sqrt{n-1}}{\sqrt{n}}\right)\)
\(< \left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\)
Áp dụng vài bài toán:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)
\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+2\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+2\left(\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2005}}\right)=2-\frac{2}{\sqrt{2005}}< 2\)
Vậy \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}< 2\)
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
Chứng minh rằng \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
Chứng minh rằng: \(2\sqrt{n+1}-2\sqrt{n}< \frac{1}{\sqrt{n}}< 2\sqrt{n}-2\sqrt{n-1}\)
Từ đó suy ra: \(2004< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{1006009}}< 2005\)
bài 1: Cho \(a+b+c=0\)Chứng minh đẳng thức
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
Bài 2: Cho \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)
\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2005}}\)
a, Rút gọ A
b, Chứng minh \(B>2\left(\sqrt{2006}-1\right)\)
Giúp mk vs ạ !!! Cô cho bài về để ôn thi học kì mà ko pic lm nà :((( !!!
1/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow\frac{a+b+c}{abc}=0\)(đúng)
Vậy ta có ĐPCM
2/ \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2006}-\sqrt{2005}\)
\(=\sqrt{2006}-1\)
b/ Ta có
\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\)
Áp dụng vài bài toán ta có
\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2005}}\)
\(>2.\sqrt{2}-2.\sqrt{1}+2.\sqrt{3}-2.\sqrt{2}+...+2.\sqrt{2006}-2.\sqrt{2005}\)
\(=2.\sqrt{2006}-2=2\left(\sqrt{2006}-1\right)\)
a. Chứng minh: \(\sqrt{n+1}-\sqrt{n}>\frac{1}{2\sqrt{n}+1}\) (với n là số tự nhiên)
b. Chứng minh :\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2005}}< 2.\sqrt{2005}\)
a)Cho a>b>0 chứng minh rằng \(\frac{1}{a+b}\le\frac{1}{2\sqrt{ab}}\)
b) Chứng minh \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+\frac{\sqrt{4}-\sqrt{3}}{7}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}< \frac{1}{2}\)