Giải:

a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)

C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)

C = \(\frac{6}{3}.\frac{1}{18}\)

C = \(2.\frac{1}{18}\)

C = \(\frac{1}{9}\)

Vậy C = \(\frac{1}{9}\)

b) D = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

D = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)\

D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

D = \(\frac{1}{2}.\frac{2}{75}\)

D = \(\frac{1}{75}\)

Vậy D = \(\frac{1}{75}\)

c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)

E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)

E = \(\frac{1}{8}-\frac{1}{41}\)

E = \(\frac{33}{328}\)

Vậy E = \(\frac{33}{328}\)

B=1/25.27+1/27.29+1/29.31+.......+1/73.75

=1/25+1/75

=3/75+1/75

=4/75.

#)Giải :

\(B=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(2B=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)

\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2B=\frac{1}{25}-\frac{1}{75}\)

\(2B=\frac{2}{75}\)

\(B=\frac{2}{75}\div2\)

\(B=\frac{1}{75}\)

B = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

   = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)

   = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

   = \(\frac{1}{2}.\frac{2}{75}\)

   = \(\frac{1}{75}\)

\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)

x=0.78

\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)

\(=3\cdot\frac{23}{200}\)

đúng

Đặt 3 ra ngoài

\(\Rightarrow B=3\left(\frac{3}{8.11}\right)+3\left(\frac{3}{11.14}\right)+..+3\left(\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{200}\right)=3.\frac{3}{25}=\frac{9}{25}\)

Vậy \(B=\frac{9}{25}\)

Chúc bn học tốt..!

\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)

\(\Rightarrow2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)

\(\Rightarrow2A=\frac{1}{25}-\frac{1}{75}=\frac{3}{75}-\frac{1}{75}=\frac{2}{75}\)

\(\Rightarrow A=\frac{2}{75}\div2=\frac{1}{75}\)

B = \(\frac{1}{25}\)- \(\frac{1}{27}\)+ \(\frac{1}{27}\)-\(\frac{1}{29}\)+\(\frac{1}{29}\)-\(\frac{1}{31}\)+... + \(\frac{1}{73}\)-\(\frac{1}{75}\)=

B = \(\frac{1}{25}\)-\(\frac{1}{75}\)

B = \(\frac{2}{75}\)

Ủng hộ  mik nha, mk đg âm điểm nè

B = 1/2.(1/25-1/27+1/27-1/29+....+1/73-1/75)

    = 1/2.(1/25-1/75)

   =1/2.2/75

   = 1/75

kik cho mk nhé. đúng đấy. kb luôn

4/75 nha bn

\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

=\(\frac{3}{20}=0,15\)

\(=1\div3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)

Bài 1:

a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=1008\cdot\left(1-\frac{1}{2017}\right)\)

Bài 2:

a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{2}{7}\)

b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)

\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)

\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(=\frac{5}{3}\cdot\frac{6}{28}\)

\(=\frac{15}{14}\)

Bài 3:

a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)

\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)

\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)

\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)

\(=-\left[10\cdot\frac{4}{55}\right]\)

\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)

\(\Leftrightarrow x=\frac{94}{99}\)

b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

\(x+1=18\)

\(x=17\)

bài dài nên tôi ko viết lại đề đâu nhé