Hh

P = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  ... + 1/8 - 1/9

= 1/2 - 1/9

= 7/18

P = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  ... + 1/8 - 1/9

= 1/2 - 1/9

= 7/18.

A= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/ 7.8 + 1/ 8.9 + 1/ 9.10 

=> A = 1/1 - 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 + 1/8 - 1/9 + 1/9 - 1/10

=> A = 1 - 1/10 = 9/10

Vậy A = 9/10

A = 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 + 1/8 - 1/9 + 1/9 - 1/10

A = 1 - 1/10 = 9/10

A= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/ 7.8 + 1/ 8.9 + 1/ 9.10

= 1/1 - 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 + 1/8 - 1/9 + 1/9 - 1/10

= 1 - 1/10 = 9/10

B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B =  \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\)) 

B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\) 

B = - \(\dfrac{1}{15}\)

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9

các câu 2;3 còn lại giống câu 1 bạn nhé

bạn thay số vào rồi làm tương tự

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9.

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

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

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

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

Vậy, \(C=\dfrac{99}{100}\)

=12/13 may ban oi lam loi giai ra cho minh vui

mk xin lỗi mk mới hok lớp 5

A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/.910+1/10.11+1/12.13

  = 1- 1/2 + 1/2 - 1/3 + ......+1/12 - 1/13

  = 1- 1/13

  = 12/13

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=2-1/1.2+3-2/2.3+4-3/3.4+...+10-9/9.10

=1-1/2+1/2-1/3+1/3-1/4+....+1/9-1/10

=1-1/10

=9/10

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

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

\(B=\frac{1}{3}-\frac{1}{10}\)

\(B=\frac{7}{30}\)

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