\(\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{82\cdot87}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HA
\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}\)
1
10 tháng 8 2017
\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
NT
\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{8}{20\cdot28}\)
2
14 tháng 7 2017
\(\frac{4}{3\cdot7}+\frac{5}{7.12}+..+\frac{8}{20\cdot28}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}+0+...+0\)
\(=\frac{25}{84}\)
14 tháng 7 2017
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}\)
\(=\frac{25}{84}\)
29 tháng 7 2018
\(\frac{2}{5\cdot7}+\frac{5}{7\cdot12}+\frac{8}{12\cdot20}+\frac{3}{140}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{20}+\frac{3}{140}\)
\(=\frac{1}{5}-\frac{1}{20}+\frac{3}{140}\)
\(=\frac{3}{20}+\frac{3}{140}=\frac{6}{35}\)
29 tháng 7 2018
\(\frac{2}{5.7}\)+ \(\frac{5}{7.12}\)+ \(\frac{8}{12.20}\)+ \(\frac{3}{140}\).
= \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{12}\)+ \(\frac{1}{12}\)- \(\frac{1}{20}\)+ \(\frac{3}{140}\).
= \(\frac{1}{5}\)- \(\frac{1}{20}\)+ \(\frac{3}{140}\).
= \(\frac{20}{100}\)- \(\frac{5}{100}\)+ \(\frac{3}{140}\).
= \(\frac{15}{100}\)+ \(\frac{3}{140}\).
= \(\frac{3}{20}\)+ \(\frac{3}{140}\).
= \(\frac{21}{140}\)+ \(\frac{3}{140}\).
= \(\frac{24}{140}\).
= \(\frac{6}{35}\).
AK
7 tháng 5 2018
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{11}{11}-\frac{1}{11}\)
\(=\frac{10}{11}\)
Chúc bạn học tốt !!!
26 tháng 3 2018
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
=\(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
=\(2\left(1-\frac{1}{21}\right)\)
=\(\frac{2.20}{21}=\frac{40}{21}\)
1 tháng 3 2019
a) \(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(C=1.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(C=1.\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(C=1.\frac{9}{38}\)
\(C=\frac{9}{38}\)
b) \(D=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{99.100}\)
\(D=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\right)\)
\(D=5.\left(\frac{1}{10}-\frac{1}{100}\right)\)
\(D=5.\frac{9}{100}\)
\(D=\frac{99}{20}\)
3 tháng 5 2018
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
Ta có:
\(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{82.87}\)
\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{82}-\frac{1}{87}\right)\)
\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{87}\right)=\frac{3}{5}.\frac{85}{174}=\frac{17}{58}\)
Ta có:
A = \(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{82.87}\)( dấu chấm là dấu nhân nhá, lên lớp 6 thì bạn sẽ biết )
\(\frac{5}{3}.A=\frac{5}{3}.\left(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{82.87}\right)\)
\(\frac{5}{3}.A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{82.87}\)
\(\frac{5}{3}.A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}+\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{82}-\frac{1}{87}\)
\(\frac{5}{3}.A=\frac{1}{2}-\frac{1}{87}\)
\(\frac{5}{3}.A=\frac{85}{174}\)
\(A=\frac{85}{174}:\frac{5}{3}=\frac{17}{58}\)
Đ/s : \(\frac{17}{58}\)
Ủng hộ mik nhá !!!