\(\dfrac{1}{1\cdot2}>\dfrac{1}{2^2}>\dfrac{1}{2\cdot3},\dfrac{1}{2\cdot3}>\dfrac{1}{3^2}>\dfrac{1}{3\cdot4},...,\dfrac{1}{8\cdot9}>\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(\Rightarrow1-\dfrac{1}{9}>A>\dfrac{1}{2}-\dfrac{1}{10}\) \(\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\)

Ta có :

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)

....................

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+........+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\left(1\right)\)

Lại có :

\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)

\(\dfrac{1}{6^2}>\dfrac{1}{6.7}\)

............

\(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)

\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+......+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+.....+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+.....+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Đặt: \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)

Ta có: \(\left\{{}\begin{matrix}A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\\A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\end{matrix}\right.\)

Vậy \(\dfrac{1}{6}< A< \dfrac{1}{4}\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)

Ta có ; 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

...

\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2012.2013}\)

\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\)

\(\Leftrightarrow B< 1-\dfrac{1}{2013}\)

\(\Rightarrow B< \dfrac{2012}{2013}\)

Lại có : \(\dfrac{2012}{2013}< \dfrac{3}{4}\)

\(\Rightarrow B< \dfrac{3}{4}\)

* Chắc vậy, sai thì thôg cảm ^^ * 

Còn j k hiểu thì ib nha

Ta thấy 3²-2²=5; 4²-3²=7;......;2014²-2013²=(2014-2013)(2014+2013)=4017

=> VT=1/4+1/4-1/9+1/9-1/16+1/16-......-1/2013²+1/2013²-1/2014²

=1/4+1/4-1/2014=1/2-1/2014²<1/2<1

\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)

\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)

\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)

\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)

\(2A=2+3+4+5+6+...+2012+2013+2014\)

\(2A=\dfrac{\left(2+2014\right).2013}{2}\)

\(A=\dfrac{2016.2013}{4}=504.2013\)

\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)

\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)

\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)

\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)

\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)

\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)

\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)

\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)