Tính các tổng sau
A = 3/4.7 + 4/7.11 + 4/11.15 + ... + 4/100.104
B = 1/25.27 + 1/27.29 + 1/29.31 +...+ 1/73.75
C = 6/15.18 + 6/18.21 + 6/21.24 +...+6/87.90
Các ban ơi dấu chấm là dấu nhân nhé nhớ giải giúp mình nhé mình thank you các bạn nhiều lắm
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Những câu hỏi liên quan
21 tháng 2 2016
=> A = 6/3.( 1/15 - 1/18 + 1/18 - 1/21 + ..... + 1/87 - 1/90 )
=> A = 2.( 1/15 - 1/90 )
=> A = 2.5/90
=> A = 10/90 = 1/9
21 tháng 1 2017
Giải:
a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\frac{1}{18}\)
C = \(2.\frac{1}{18}\)
C = \(\frac{1}{9}\)
Vậy C = \(\frac{1}{9}\)
b) D = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
D = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)\
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\frac{2}{75}\)
D = \(\frac{1}{75}\)
Vậy D = \(\frac{1}{75}\)
c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)
E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)
E = \(\frac{1}{8}-\frac{1}{41}\)
E = \(\frac{33}{328}\)
Vậy E = \(\frac{33}{328}\)
26 tháng 2 2022
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)
29 tháng 3 2016
A = 6/3 . ( 1/15.18 + 1/18.21 + 1/21/24 + . . . + 1/87.90 )
A = 6/3 . ( 1/15 - 1/18 + 1/18 - 1/21 + 1/21 - 1/24 + . . . + 1/87 - 1/90 )
A = 2 . ( 1/15 - 1/90 )
A = 2. 5/90
A = 10/90 = 1/9
29 tháng 3 2016
\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{84.87}+\frac{6}{87.90}\)
\(=\frac{6}{3}\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{84.87}+\frac{3}{87.90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{84}-\frac{1}{87}+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{90}\right)=2\left(\frac{6-1}{90}\right)=2\times\frac{1}{18}=\frac{1}{9}\)
TT
1
5 tháng 8 2016
Ta có:
\(A=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(A=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(A=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(A=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(A=2.\frac{1}{18}=\frac{1}{9}\)
NH
7 tháng 8 2021
Ta có :
\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(2B=\frac{2}{25.27}+\frac{1}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)
\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+....+\frac{1}{73}-\frac{1}{75}\)
\(2B=\frac{1}{25}-\frac{1}{75}\)
\(2B=\frac{2}{75}\)
\(\Rightarrow B=\frac{1}{75}\)
Vậy B = \(\frac{1}{75}\)
NH
7 tháng 8 2021
\(F=\frac{4}{2.3}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(\Rightarrow F=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(\Rightarrow F=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Rightarrow F=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(\Rightarrow F=2.\frac{502}{1005}=\frac{1004}{1005}\)
Vậy F = \(\frac{1004}{1005}\)
YN
4 tháng 6 2021
\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(\rightarrow\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(\rightarrow\frac{1}{9}\)
TK
3
10 tháng 7 2018
a, \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}\left(\frac{2}{75}\right)\)
\(=\frac{1}{75}\)
b, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1004}{2010}\right)\)
\(=2\left(\frac{502}{1005}\right)\)
\(=\frac{1004}{1005}\)
Tk hộ =v
10 tháng 7 2018
\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}.\frac{2}{75}=\frac{1}{75}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)
\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)
\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)
\(A=\frac{1}{4}-\frac{1}{104}\)
\(A=\frac{25}{104}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)
\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)
\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)
\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)
\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)
\(B\cdot2=\frac{2}{75}\)
\(B=\frac{2}{75}\frac{\cdot}{\cdot}2\)
\(B=\frac{1}{75}\)