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\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{7\cdot9}+\dfrac{1}{6\cdot8}\)
\(=\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{7\cdot9}\right)+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{7\cdot9}\right)+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{9}{9}-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}+\dfrac{1}{2}\cdot\dfrac{3}{8}\)
\(=\dfrac{1}{2}\left(\dfrac{8}{9}+\dfrac{3}{8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{64}{72}+\dfrac{27}{72}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{91}{72}\)
\(=\dfrac{91}{144}\)
S=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+...+\dfrac{1}{6.8}\)
S=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{2}+...+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
S=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{8}\right)\)
S=\(\dfrac{1}{2}.\left(\dfrac{8-1}{8}\right)\)
S=\(\dfrac{1}{2}.\dfrac{7}{8}\)
S=\(\dfrac{7}{16}\)
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
2
a. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{2}-\dfrac{1}{100}\)
=\(\dfrac{49}{100}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-1}{2x}=\dfrac{-1}{4}\)
\(\Rightarrow x=2\)
Ta có: \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\Rightarrow\dfrac{1}{2}.\dfrac{x-1}{2x}=\dfrac{1}{8}\Rightarrow\dfrac{x-1}{4x}=\dfrac{1}{8}\)
\(\Rightarrow8\left(x-1\right)=4x\Rightarrow8x-8=4x\Rightarrow4x=8\Rightarrow x=2\)
Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)
A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)
A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
A=\(\dfrac{98}{99}-\dfrac{49}{100}\)
A=\(\dfrac{4949}{9900}\)
Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)
Vậy A<\(\dfrac{3}{4}\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
A=\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{20\cdot22}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{11}{22}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{5}{11}\)
\(=\dfrac{5}{22}\)
Đặt \(A=\dfrac{1}{4.6}+\dfrac{1}{2.6.8}+\dfrac{1}{3.8.10}+\dfrac{1}{4.10.12}\)
\(\dfrac{1}{2}A=\dfrac{1}{2.4.6}+\dfrac{1}{4.6.8}+\dfrac{1}{6.8.10}+\dfrac{1}{8.10.12}\)
\(\left(\dfrac{1}{2}A\right).4=\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\)
\(\dfrac{1}{2}.4.A=\dfrac{6-2}{2.4.6}+\dfrac{8-4}{4.6.8}+\dfrac{10-6}{6.8.10}+\dfrac{12-8}{8.10.12}\)
\(2A=\dfrac{6}{2.4.6}-\dfrac{2}{2.4.6}+\dfrac{8}{4.6.8}-\dfrac{4}{4.6.8}+\dfrac{10}{6.8.10}-\dfrac{6}{6.8.10}+\dfrac{12}{8.10.12}-\dfrac{8}{8.10.12}\)
\(2A=\dfrac{1}{2.4}-\dfrac{1}{4.6}+\dfrac{1}{4.6}-\dfrac{1}{6.8}+\dfrac{1}{6.8}-\dfrac{1}{8.10}+\dfrac{1}{8.10}-\dfrac{1}{10.12}\)
\(2A=\dfrac{1}{2.4}-\dfrac{1}{10.12}\)
\(2A=\dfrac{1}{8}-\dfrac{1}{120}\)
\(2A=\dfrac{7}{60}\)
\(A=\dfrac{7}{120}\)