LN
6
K
Khách
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NT
3
KT
2 tháng 8 2018
\(A=\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+.....+\frac{2}{73.76}\)
\(=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{73.76}\right)\)
\(=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{73}-\frac{1}{76}\right)\)
\(=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(=\frac{2}{3}.\frac{9}{38}=\frac{3}{19}\)
CD
2
DL
29 tháng 6 2017
Đặt : \(A=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{27\cdot30}\)
\(A=\frac{1}{3}\left(\frac{5}{1}-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+...+\frac{5}{27}-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\left(5-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\cdot\frac{29}{6}\)
\(A=\frac{29}{18}\)
DP
29 tháng 6 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+....+\frac{5}{27.30}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{30-27}{27.30}\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\frac{29}{30}=\frac{29}{18}\)
VN
Võ Ngọc Phương
CTVHS
VIP
21 tháng 8 2023
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)
\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=\dfrac{1}{103}\)
\(\Rightarrow x+3=103\)
\(x=103-3\)
\(x=100\)
Vậy x = 100
BD
3
19 tháng 3 2018
Ta có :
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
Vậy \(A=\frac{33}{50}\)
Chúc bạn học tốt ~
T
19 tháng 3 2018
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
18 tháng 5 2016
\(A=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(A=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
TA
18 tháng 5 2016
A=3/10.13 +3/13.16+ 3/16.19+....+3/58.61
A=1/10.13+1/13.16+1/16.19+.....+1/58.61
A=1/10- 1/13+ 1/13- 1/16+ 1/16- 1/19+...+1/58 –1/61
A=1/10 – 1/61
A= 61/610 – 10/610
A= 51/610
Mình giải xong rồi k nhá?
29 tháng 6 2017
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
DL
1 tháng 7 2017
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)
19 tháng 7 2016
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
19 tháng 7 2016
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
Y
20 tháng 7 2019
a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
C = 3/4.7 + 3/7.10 + 3/10.13 + ... + 3/73.76
=1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/73 - 1/76
=1/4 - 1/76
=18/76
\(C=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+......+\frac{1}{73}-\frac{1}{76}\)
\(=\frac{1}{4}-\frac{1}{76}\)
\(=\frac{19}{76}-\frac{1}{76}\)
\(=\frac{18}{76}=\frac{9}{38}\)