a, \(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{2011\cdot2011}\)

có :

\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{2011\cdot2011}< \frac{1}{2010\cdot2011}\)

nên :

\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2010\cdot2011}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(\Rightarrow A< 1-\frac{1}{2011}\)

\(\Rightarrow A< \frac{2010}{2011}< 1\)

b, \(A=\frac{2010}{2011}=1-\frac{1}{2011}\) 

\(\frac{3}{4}=1-\frac{1}{4}\)

\(\frac{1}{4}>\frac{1}{2011}\)

nên :

\(A>\frac{3}{4}\)

a, A bé hơn 1

b, A bé hơn 3/4

hello mây chế

Ta thấy:

1/2*2<1/1*2)vì 2*2>1*2).

1/3*3<1/2*3(vì 3*3>2*3).

...

1/8*8<1/7*8(vì 8*8>7*8).

=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.

=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.

=>B<1-1/8.

=>B<7/8.

Mà 7/8<1.

=>B<1.

Vậy B<1(đpcm).

\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(\Rightarrow1-\frac{1}{8}< 1\)

=>B<1

\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6.}+\frac{1}{7.7}+\frac{1}{8.8}\)\(=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(B=1-\frac{1}{8}\)

\(\Rightarrow B< 1\left(ĐPCM\right)\)

 \(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+....+\frac{1}{10\cdot10}\)

Ta có : 

\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

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

\(\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)

Ta có : 

\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1}-\frac{1}{10}\)

\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{9}{10}\)

\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{9}{10}< 1\)

Đặt \(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow B< 1-\frac{1}{10}< 1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}\)

\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(...\)

\(\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)

Cộng vế theo vế 

\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< 1-\frac{1}{10}\)

Lại có \(1-\frac{1}{10}< 1\)

\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< 1-\frac{1}{10}< 1\)

\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< 1\)( đpcm )

help me!!!

                                                                     \(Giải\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)

      \(A=0+0+0+...+0+0\)

      \(\Rightarrow A=0\)   

\(a.\)\(A< 1\)

b.   \(A< \frac{3}{4}\)

Ta có: \(S=1+\frac{1}{2x2}+\frac{1}{3x3}+.....+\frac{1}{10x10}\)

Ta có: 1/2x2 < 1/1x2

          1/3x3 < 1/2x3 

          1/4x4 < 1/3x4

         .......................

         1/10x10 < 1/9x10

=> S< 1+1/1x2+1/2x3+1/3x4+.....+1/9x10

=> S<1+(1-1/10)

=> S < 1+9/10

=> S < 19/10 < 2

Vậy S<2 

1/5x5;1/6x6;1/7x7;1/8x8;1/9x9

đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(B=1-\frac{1}{10}<1\) (1)

Mà 1<2 (2)

Ta có:\(S=\frac{1}{1.1}+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{10.10}(3)

ta co 

1/2.2<1/1*2

...

1/2018*2018<1/2017*2018

=>1/2*2+...+1/2018*1018<1/1*2+...+1/2017.2018

.....(tinh 1/1*2+...+1/2017.*2018)

=>1/2*2+...+1/2018*2018<1-1/2018<1

=>1/2*2+...+1/2018*2018<1

\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)

\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(A>1-\frac{1}{2015}\)

Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)

1.

\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{2012}{2013}\)

\(A=\frac{1.2.3.4.....2012}{2.3.4.5......2013}\)

\(A=\frac{1}{2013}\)

\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)

\(B=\frac{2012\left(2013-2012\right)}{2012\left(2011+2\right)}\)

\(B=\frac{2013-2012}{2011+2}\)

\(B=\frac{1}{2013}\)

\(Vì:\frac{ 1}{2013}=\frac{1}{2013}\)

\(\Rightarrow\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)

\(Hay: A=B\)

\(A=\frac{1\times2}{2\times2}\times\frac{2\times3}{3\times3}\times\frac{3\times4}{4\times4}\times\frac{4\times5}{5\times5}\times...\times\frac{2012\times2013}{2013\times2013}\)

\(\Rightarrow A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2012}{2013}\)

\(\Rightarrow A=\frac{1\times2\times3\times4\times...\times2012}{2\times3\times4\times5\times...\times2013}\)

\(\Rightarrow A=\frac{1}{2013}\)

\(B=\frac{2012\times2013-2012\times2012}{2012\times2011+2012\times2}\)

\(\Rightarrow B=\frac{2012\times\left(2013-2012\right)}{2012\times\left(2011+2\right)}\)

\(\Rightarrow B=\frac{2012\times1}{2012\times2013}\)

\(\Rightarrow B=\frac{1}{2013}\)