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

D=2.\(\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

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

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

Vậy D=\(^{\frac{1}{9}}\)

\(D=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)

\(D=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)

\(D=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(D=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(D=2.\left(\frac{6}{90}-\frac{1}{90}\right)\)

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

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

Ta có: \(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)

\(=2\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)

\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(=2\cdot\frac{1}{18}=\frac{1}{9}\)

\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+.......+\frac{6}{87.90}\)

\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+.......+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)

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

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

\(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\)

\(=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{84}-\frac{1}{87}+\frac{1}{87}-\frac{1}{90}\)

\(=\frac{1}{15}-\frac{1}{90}\)

\(=\frac{6}{90}-\frac{1}{90}\)

\(=\frac{5}{90}\)

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

1/15-1/18+1/18-1/21+1/21-1/24+....+1/87-1/90

=1/15-1/90

=6/90-1/90

=5/90

=1/16

\(=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\)

\(=\frac{1}{15}-\frac{1}{90}=\frac{1}{18}\)

A = 6/3 . ( 1/15.18 + 1/18.21 + 1/21/24 + . . . + 1/87.90 )

A = 6/3 . ( 1/15 - 1/18 + 1/18 - 1/21 + 1/21 - 1/24 + . . . + 1/87 - 1/90 )

A = 2 . ( 1/15 - 1/90 ) 

A = 2. 5/90

A = 10/90 = 1/9

\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{84.87}+\frac{6}{87.90}\)

\(=\frac{6}{3}\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{84.87}+\frac{3}{87.90}\right)\)

\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{84}-\frac{1}{87}+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2\left(\frac{1}{15}-\frac{1}{90}\right)=2\left(\frac{6-1}{90}\right)=2\times\frac{1}{18}=\frac{1}{9}\)

\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)

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

\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(\rightarrow\frac{1}{9}\)

=> A = 6/3.( 1/15 - 1/18 + 1/18 - 1/21 + ..... + 1/87 - 1/90 )

=> A = 2.( 1/15 - 1/90 )

=> A = 2.5/90

=> A = 10/90 = 1/9

gfdddddddddddddrh

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

C = \(2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

C = \(2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\frac{1}{18}\)

C = \(\frac{1}{9}\)

\(B=\frac{6}{1.3}+\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{99.101}\)

\(=3.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{9}{99.101}\right)\)

\(=3.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=3.\left(\frac{1}{1}-\frac{1}{101}\right)=3.\left(\frac{101}{101}-\frac{1}{101}\right)=3.\frac{100}{101}=\frac{300}{101}\)

\(C=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)

\(=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)

\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+....+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\left(\frac{6}{90}-\frac{1}{90}\right)=2.\frac{5}{90}=\frac{1}{9}\)

a,A=\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{23.24}\)

A=\(\frac{1}{2}+\frac{2}{1}-\frac{1}{3}+\frac{3}{1}-\frac{1}{4}+......\frac{23}{1}-\frac{1}{24}\)

A=\(\frac{1}{2}-\frac{1}{24}\)

A=\(\frac{11}{24}\)

Còn câu b bạn??

b)\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

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

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

\(=3.\frac{5}{90}\)

\(=\frac{5}{30}\)

\(=\frac{1}{6}\)