\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)

\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(A=2.\frac{12}{25}=\frac{2.12}{25}=\frac{24}{25}\)

cảm ơn bạn . kb với mik nha

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

=\(1-\frac{1}{50}\)

Vì \(1-\frac{1}{50}< 1\)nên A < 1

B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=\(\frac{1}{2}-\frac{1}{100}\)

Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(\Rightarrow A< 1\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=\frac{1}{2}-\frac{1}{100}\)

\(\Rightarrow B< \frac{1}{2}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}< 1\)

\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

3A = 49.50.51

A = 41650

Thay vào ta được

41650 + 1/2x = 40642

=> 1/2x = 1008

=> x = 2016

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=2.\frac{49}{100}\)

\(=\frac{49}{50}\)

= 2.(1/2.3 + 1/3.4 + ... + 1/99.100)

trong ngoac co cong thuc do, tim hieu di la lam dc

\(2\cdot A=2\cdot\left(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{99\cdot100}\right)\)

\(2\cdot A=\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+\frac{4}{4\cdot5}+...+\frac{4}{99\cdot100}\)

\(2\cdot A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(2\cdot A=1-\frac{1}{100}\)
\(2\cdot A=\frac{99}{100}\)

\(A=\frac{99}{100}:2=\frac{99}{200}\)

ta có : 1/1.2+1/2.3+1/3.4+1/4.5+....+1/49.50

= 1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.....+1/49-1/50

=1/1-1/50

= 49/50

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{1}-\frac{1}{50}\)

\(=\frac{49}{50}\)

ta có : 1/1.2+1/2.3+1/3.4+1/4.5+....+1/49.50

= 1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+....+1/49-1/50

=1/1-1/50

=49/50

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)

\(=2.\left(1-\frac{1}{20}\right)\)

\(=2.\frac{19}{20}=\frac{19}{10}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{1}-\frac{1}{20}\)

\(=\frac{19}{20}\)

Đây là toán lớp 6 mà bạn. Lớp 5 ở đâu ra có dạng bài toán này

dấu bằng của mk bt liệt nên bạn thông cảm 

A bằng (1.2.3.4).(1.2.3.4)/(1.2.3.4).(2.3.4.5) bằng 5 

rút gọn cho nhau bạn nhé

thank bạn nhóa

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

\(\frac{2}{1}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(\frac{2}{1}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\frac{2}{1}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\frac{2}{1}.\frac{49}{100}\)

\(\frac{98}{100}=\frac{49}{50}\)

Đặt A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

 A : 2 =  \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

 A : 2 = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

 A : 2 = \(\frac{1}{2}-\frac{1}{100}\)

 A : 2 = \(\frac{49}{100}\)

    A   = \(\frac{49}{50}\)

Đặt \(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)

\(\Rightarrow A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\Rightarrow A=2.\frac{49}{50}\)

\(\Rightarrow A=\frac{49}{25}\)

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{1}{5}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)