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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow M=6\left(\frac{1}{15.18}+\frac{1}{18.21}+\frac{1}{21.24}+....+\frac{1}{87.90}\right)\)

\(\Rightarrow M=6\left[\frac{1}{3}\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+.....+\frac{1}{87}-\frac{1}{90}\right)\right]\)

\(\Rightarrow M=6\left[\frac{1}{3}\left(\frac{1}{15}-\frac{1}{90}\right)\right]\Rightarrow M=6\left(\frac{1}{3}.\frac{1}{18}\right)\Rightarrow M=6.\frac{1}{54}\Rightarrow M=\frac{1}{9}\)

Kết quả là : \(\frac{2}{18}\)

   Ngại viết quá ! k nha!

  ^_^ 

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??

Ta có: \(A=\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\)

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

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

                \(=\frac{1}{3}(\frac{6}{90}-\frac{1}{90})\)

                \(=\frac{1}{3}.\frac{5}{90}\)

                \(=\frac{1}{54}\)

Ta có: 1= \(\frac{54}{54}\)

Suy ra A < 1 (đpcm)

3A=3*(1/15*18+1/18*21+...+1/87*90)

3A=3/15*18+3/18*21+...+3/87*90

3A=1/15-1/18+1/18-1/21+...+1/87-1/90

3A=1/15-1/90

3A=1/18

A=1/18 chia3

A=1/54

vì 1/54<1 nên A<1

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

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

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

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

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

A = 6/3 . ( 1/15.18 + 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

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

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

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

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