`@` `\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, \(C=\dfrac{99}{100}\)

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=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

A=1-1/100                            A=99/100                                                                                    B= (1/5.6+1/6/7+...+1/101.102).3                         B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3        B=(1/5-1/102).3                                                 B=97/170                                                            

1) Tính

a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

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

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

3.(1/4.5+1/5.6+...+1/9.10).x=9/2

3.(1/4-1/5+1/5-1/6+...+1/9-1/10).x=9/2

3.(1/4-1/10).x=9/2

3.3/20.x=9/2

9/20.x=9/2

x=9/2:9/20

x=10

Ta có:

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

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

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

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

\(=\dfrac{1}{100}\)

2/ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+.........+\dfrac{1}{99.100}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+........+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{5}-\dfrac{1}{100}\)

\(=\dfrac{19}{100}\)

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

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

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{98}{99}.\dfrac{99}{100}\)

\(=\dfrac{1}{100}\)

2. \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{5}-\dfrac{1}{100}\)

\(=\dfrac{20}{100}\) \(-\dfrac{1}{100}\)

\(=\dfrac{19}{100}\)

đây là tính nhanh à nếu tính bình thường thì tính may tính là ra

a) 17/23 . 8/16 . 23/17. (-80) . 3/4

= (17/23 . 23/17) . (8/16 . 3/4) . (-80)

= 1 . 3/8 . (-80)

= 3/8 . (-80)

= -30

b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29

= 5/11 . (18/29 - 8/29 + 19/29)

= 5/11 . 1

= 5/11

c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)

= (13/23 + 1313/2323 - 131313/232323).0

= 0

d) 1²/2x2 . 2²/2x3 . 3²/3x4 . 4²/4x5 . 5²/5x6 . 6²/6x7 . 7²/7x8 . 8²/8x9 . 9²/9x10

= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10

= 1/10

Khó nhìn quá. Bạn thông cảm nhé!

a)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

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

\(=\dfrac{4}{25}\)

b)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)

tương tự

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)

\(A=\dfrac{101+102+103+...+109}{100}\)

Xét tử số : \(101+102+103+...+199\)

Có : \(\left(199-101\right):1+1=99\) (số hạng)

\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)

\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)

\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

Xét tử số : \(10002+10004+10006+...+12014\)

Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)

\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)

\(\Rightarrow B=\dfrac{11085056}{10000}\)

Bạn tự làm câu C nha

\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)

\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)

\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)

\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)

\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)

D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)

E = \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}\)+ ...+ \(\dfrac{1}{2014.2017}\)
E = \(\dfrac{1.3}{3.1.4}+\dfrac{1.3}{3.4.7}+\dfrac{1.3}{3.7.10}+...+\dfrac{1.3}{3.2014.2017}\)
E = \(\dfrac{1}{3}\) .( \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\) )
E = \(\dfrac{1}{3}\).\(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\left(1-\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\left(\dfrac{2015}{2015}-\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\dfrac{2014}{2015}\)
E = \(\dfrac{2014}{6045}\)