\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)

\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)

\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

\(2S=\frac{1}{2}-\frac{1}{10}\)

\(2S=\frac{2}{5}\)

\(S=\frac{2}{5}:2\)

\(S=\frac{1}{5}\)

S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(\Rightarrow2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(\Rightarrow2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

\(\Rightarrow2S=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S=\frac{2}{5}:2=\frac{2}{5}.\frac{1}{2}=\frac{1}{5}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

  \(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(=\frac{4}{9}-\frac{1}{5}\)

\(=\frac{11}{45}\)

\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{4}{9}-\frac{1}{5}\)

\(S=\frac{11}{45}\)

\(\frac{x}{2^2}\)+\(\frac{x}{2^3}\) +\(\frac{x}{2^4}\) =\(\frac{x}{3^2}\) +\(\frac{x}{3^3}\) +\(\frac{x}{3^4}\) là x =

=>2A=2(1/2x4+1/4.6+1/6.8+1/8.10+1/10.12+1/12.14)

=> 2A=2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 + 2/10.12 + 2/12.14

=> 2a =1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7

=> 2A =1-1/7

=>2A=16/17

=> A= 8/17

Mình chắc chắn . Chúc bạn học tốt

\(A=\frac{1}{2.4}\)\(+\frac{1}{4.6}\)\(+\frac{1}{6.8}\)\(+\frac{1}{8.10}\)\(+\frac{1}{10.12}\)\(+\frac{1}{12.14}\)

\(\Rightarrow2A=2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{14}=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}\)

\(\Rightarrow2A=\frac{6}{14}\)

\(\Rightarrow A=\frac{3}{14}\)

=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)

= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)

= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)

=\(\frac{29}{45}\)

29/45 bạn nhé

<br class="Apple-interchange-newline"><div id="inner-editor"></div>S=11.3 −12.4 +13.5 −14.6 +15.7 −16.8 +17.9 −18.10 

(11.3 +13.5 +15.7 +17.9 )−12.4 −14.6 −16.8 −18.10 

=12 (21.3 +23.5 +25.7 +27.9 )−(12.4 +14.6 +16.8 +18.10 )

=12 (21.3 +23.5 +25.7 +27.9 )−12 (22.4 +24.6 +26.8 +28.10 )

=12 (1−13 +13 −15 +15 −17 +17 −19 )−12 (12 −14 +14 −16 +16 −18 +18 −110 )

=12 (1−19 )−12 (12 −110 )

=12 (1−19 −12 +110 )

a) Ta có:

\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)

\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)

\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)

\(=x+2x+-3+1-21\)

\(=3x-23\)

=> \(3x-23=2020\)

\(3x=2020+23=2043\)

=> \(x=2043:3=681\)

Nhầm

\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)

\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)

Ta có

=\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{8.10}\right)\)

=\(\frac{4}{3}.\frac{9}{8}....\frac{81}{80}\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{9.9}{8.10}\)

=\(\frac{2.3....9}{1.2....8}.\frac{2.3....9}{3.4....10}\)

=\(9.\frac{2}{10}\)

=\(\frac{9}{5}\)