A= \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{100}=\frac{99}{100}\)

=> A= \(\frac{99}{100}>\frac{25}{26}\)

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

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

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

Cách này khó hiểu quá

\(C=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{100\times100}\\ C< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ C< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ C< 1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)

 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

giúp mình nha

Ta có : A = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

\(\Rightarrow A< \frac{8}{9}\)(1)

Lại có : \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow A>\frac{2}{5}\)(2)

Từ (1);(2) => \(\frac{8}{9}>A>\frac{2}{5}\)

cảm ơn nha. bài này mình đang cần gấp nè

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

\(A=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+...+\dfrac{1}{100\times100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Quy đồng 99/100 với 3/4, ta có:

\(\dfrac{99}{100}=\dfrac{396}{400};\dfrac{3}{4}=\dfrac{300}{400}\)

So sánh A với 3/4: \(\dfrac{99}{100}>\dfrac{3}{4}\left(\dfrac{396}{400}>\dfrac{300}{400}\right)\)

2.2+3.3+4.4+...+100.100

= 2²+3²+4²+...+100²

= 1²+2²+3²+...+100²-1

=\(\dfrac{100.\left(100+1\right).\left(2.100+1\right)}{6}\)-1

=338350-1

=338349