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27 tháng 4 2017

\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)

\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

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

\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)

21 tháng 6 2017

\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

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

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

21 tháng 6 2017

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)

\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

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

Vậy \(S=\dfrac{99}{100}\)

Chúc bạn học tốt!!!

11 tháng 4 2022

\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)

=> \(\dfrac{2.3}{1.4}+\dfrac{2.3}{4.7}+...+\dfrac{2.3}{97.100}\)

=> \(2.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

=> \(2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

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

=>\(2\).\(\dfrac{99}{100}\)

=\(\dfrac{99}{50}\)

29 tháng 4 2022

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

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

\(2.\dfrac{99}{100}\)

=\(\dfrac{99}{50}\)

\(=\dfrac{5^3\cdot2^3+2\cdot5^3+5^3}{55}=\dfrac{5^3\left(2^3+2+1\right)}{55}=5^2=25\)

11 tháng 2 2023

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`

`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`

`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`

`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`

`S_1 = 5/3 (1 - 1/100)`

`S_1 = 5/3 . 99/100`

`S_1 = 33/20`

12 tháng 4 2022

\(2.5+1.4-1\dfrac{3}{7}=\dfrac{5}{2}+\dfrac{7}{5}-\dfrac{10}{7}=\dfrac{5\cdot35}{2\cdot35}+\dfrac{7\cdot14}{5\cdot14}-\dfrac{10\cdot10}{7\cdot10}=\dfrac{175}{70}+\dfrac{98}{70}-\dfrac{100}{70}=\dfrac{173}{70}\)

12 tháng 4 2022

2,5 + 1,4 - \(1\dfrac{3}{7}\)

=3,9 + \(\dfrac{10}{7}\)

=\(\dfrac{373}{70}\)

3 tháng 2 2022

1.

`16 + (27 - 7.6 ) - (94 -7 - 27.99)`

`= 16+ 27 - 7.6 - 94 + 7 + 27.99`

`= 16 + 27(99 +1) - 7(6-1) - 94`

`= -78 + 27.100 - 7.5`

`= 2587`

2.

`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`

`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`

`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`

`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`

`3/2A = 1 - 1/100`

`3/2 A= 99/100`

`A= 99/100 : 3/2`

`A=33/50`

Vậy `A= 33/50`

3 tháng 2 2022

1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99

                                           =(27+27.99)+(27+7-94)+16

                                           =27.100-60+16

                                           =2700-44=2656

2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)

     =\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

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

a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)

b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)

c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)

d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

1 tháng 3 2022

đề bài là j