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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=\(\dfrac{7}{16}\)
\(D=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{98.100}\)
\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}.\dfrac{49}{100}=\dfrac{147}{200}\)
\(D=\dfrac{3}{2\cdot4}+\dfrac{3}{4\cdot6}+\dfrac{3}{6\cdot8}+...+\dfrac{3}{98\cdot100}\\ =\dfrac{3}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\dfrac{49}{100}\\ =\dfrac{147}{200}\)
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
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-1}{2x}=\dfrac{-1}{4}\)
\(\Rightarrow x=2\)
Ta có: \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\Rightarrow\dfrac{1}{2}.\dfrac{x-1}{2x}=\dfrac{1}{8}\Rightarrow\dfrac{x-1}{4x}=\dfrac{1}{8}\)
\(\Rightarrow8\left(x-1\right)=4x\Rightarrow8x-8=4x\Rightarrow4x=8\Rightarrow x=2\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}=\dfrac{9}{50}=0,18\)
Vậy \(S>\dfrac{1}{10}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}\)
\(S=\dfrac{20}{100}-\dfrac{2}{100}\)
\(S=\dfrac{18}{100}=\dfrac{9}{50}=0,18\)
\(\dfrac{1}{10}=0,1\), mà \(0,1< 0,18\)
\(\Rightarrow S>\dfrac{1}{10}\left(đpcm\right)\)
B = \(\dfrac{12}{\left(2.4\right)^2}+\dfrac{20}{\left(4.6\right)^2}+...+\dfrac{388}{\left(96.98\right)^2}+\dfrac{396}{\left(98.100\right)^2}\)
= \(\dfrac{4^2-2^2}{2^{2^{ }}.4^{2^{ }}}+\dfrac{6^{2^{ }}-4^2}{4^2.6^2}+...+\dfrac{98^2-96^2}{96^2.98^2}+\dfrac{100^2-98^2}{98^2.100^2}\)
=\(\dfrac{1}{2^{2^{ }}}-\dfrac{1}{4^{2^{ }}}+\dfrac{1}{4^2}-\dfrac{1}{6^2}+\dfrac{1}{6^2}+....-\dfrac{1}{98^2}+\dfrac{1}{98^2}-\dfrac{1}{100^2}\)
= \(\dfrac{1}{2^2}-\dfrac{1}{100^2}=\dfrac{1}{4}-\dfrac{1}{100^2}< \dfrac{1}{4}\)
Vậy B < \(\dfrac{1}{4}\)
B = 12(2.4)2+20(4.6)2+...+388(96.98)2+396(98.100)212(2.4)2+20(4.6)2+...+388(96.98)2+396(98.100)2
= 42−2222.42+62−4242.62+...+982−962962.982+1002−982982.100242−2222.42+62−4242.62+...+982−962962.982+1002−982982.1002
=122−142+142−162+162+....−1982+1982−11002122−142+142−162+162+....−1982+1982−11002
= 122−11002=14−11002<14122−11002=14−11002<14
Vậy B < 14
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<\(\dfrac{3}{4}\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
\(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
= 49/100 nha