Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt tổng trên là A ta có
\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)
\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)
\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)
\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
S=\(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+.....+\frac{2}{98.100}\)
S=\(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+........+\frac{1}{98}-\frac{1}{100}\)
S=\(\frac{1}{10}-\frac{1}{100}\)
S=\(\frac{9}{100}\)<\(\frac{1}{10}\)
=3/2(2/10.12+2/12.14+...+2/48.50)
=3/2(1/10-1/12+1/12-1/14+...+1/48-1/50)
=3/2(1/10-1/50)
=3/2 . 2/25 =3/25
Đặt \(A=\frac{3}{10.12}+\frac{3}{12.14}+.....+\frac{3}{48.50}\)
\(A=\frac{3}{2}.\left(\frac{2}{10.12}+\frac{2}{12.14}+......+\frac{2}{48.50}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{12}+....+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\frac{2}{25}\)
\(A=\frac{3}{25}\)
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}-\frac{2}{7}\)
\(=0\)
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\frac{2}{7}-\frac{2}{7}=0\)
Study well ! >_<
S=1/5.6+1/10.9+1/15.12+...+1/3350.2013
=(1/5).(1/3).(1/1.2+1/2.3+1/3.4+...+1/670.671)
=(1/15). (1-1/2+1/2-1/3+...+1/670-1/671)
=(1/15). (1-1/671)
=1/15.670/671
=134/2013
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
2/5 . A = 2/10.12 + 2/12.14 + ....... + 2/998.1000
= 1/10 - 1/12 + 1/12 - 1/14 + ..... + 1/998 - 1/1000
= 1/10 - 1/1000 = 99/1000
=> A = 99/1000 : 2/5 = 99/400
Tk mk nha
Mình làm tắt cũng dc nhé có gì ko dc hỏi mình nhé
Ta có:A=(5/10-5/12):2+(5/12-2/14):2+...+(5/998-5/1000):2
suy ra A=5/10-5/1000=99/200
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}...\frac{14^2}{13.15}\)
\(\frac{8.8}{7.9}.\frac{9.9}{8.10}...\frac{14.14}{13.15}\)
\(\frac{8.9...14}{7.8...13}.\frac{8.9...14}{9.10...15}\)
\(\frac{14}{7}.\frac{8}{15}\)
\(2.\frac{8}{15}\)
\(\frac{16}{15}\)
(8.9.10.11.12.13.14)(8.9.10.11.12.13.14)/7.8.9.10.11.12.13).(9.10.11.12.13.14.15)
=14.8/7.15
=16/15
k cho mình nhá
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)