\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}\)

\(\dfrac{1}{2.2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3.3}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4.4}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{2009.2009}< \dfrac{1}{2008.2009}=\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1\)

Ta có:

\(\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{2009\times2009}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2008\times2009}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)

 Ta có : 1/[n x (n - 1)] = [(n - 1) - n] / [n x (n - 1)] = 1/n - 1/(n - 1) 
Áp dụng : 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) 
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/48 - 1/49 + 1/49 - 1/50 
= 1 - 1/50 < 1 
Vậy : 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) < 1 
Ta có : 1/(n x n) < 1/[(n - 1) x n] 
1/(2x2) < 1/(1x2) 
1/(3x3) < 1/(2x3) 
1/(4x4) < 1/(3x4) 
............. 
1/(49x49) < 1/(49x49) 
1/(50x50) < 1/(49x50) 
=> 1/(2x2) + 1/(3x3) + 1/(4x4) + ... 1/(49x49) + 1/(50x50) < 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) < 1 
Vậy 1/(2x2) + 1/(3x3) + 1/(4x4) + ... 1/(49x49) + 1/(50x50) < 1

Đặt B=1/1*2+1/2*3+...+1/99*100 

Ta thấy:

A=1/2*2+1/3*3+...+1/100*100<B=1/1*2+1/2*3+...+1/99*100   (1)

Ta lại có: 

B=1/1*2+1/2*3+...+1/99*100 

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100<1 (2)

Từ (1) và (2) ta có: A<B<1 <=>A<1

A bé hơn 1 nha bạn

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

A = 1

nha bạn  mình chắc chắn

nhưng cách lm như têk nào hả bạn

A = 1/2×2 + 1/3×3 + 1/4×4 + ... + 1/100×100

A < 1/1×2 + 1/2×3 + 1/3×4 + ... + 1/99×100

A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

A < 1 - 1/100 < 1

\(E=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}\)

Ta có \(\frac{1}{2.2}>\frac{1}{2.3}\)

\(\frac{1}{3.3}>\frac{1}{3.4}\)

...

\(\frac{1}{49.49}>\frac{1}{49.50}\)

=> \(E=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{49.49}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}=F\)

=> E > F

Xyz olm ơi . là j vậy

help me!!!

ai nhanh nhất mình tk cho

cămmon

1, A=1/2 x (1/2 + 1/6 + 1/12+ 1/20 +1/30 +1/42)

    A= 1/2 x (1/1x2 + 1/ 2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7)

A=1/2 x (1/1 -1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6 +1/6 - 1/7 )

A= 1/2 x (1- 1/7)

A= 1/2 x 6/7

A=3/7

\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2021.2021}\)

\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2021^2}\)

Xét : \(\frac{1}{k^2}\left(k\inℕ^∗\right)\)

\(=\frac{4}{4k^2}< \frac{4}{4k^2-1}=\frac{4}{\left(2k-1\right)\left(2k+1\right)}==2\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\)

Áp dụng cho biểu thức A,ta có :

\(A< 2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{4041}-\frac{1}{4023}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{4023}\right)=\frac{2}{3}-\frac{2}{4023}< \frac{2}{3}< \frac{3}{4}\)

Yuriko

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