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

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

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

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

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

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

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

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 = 54/15.18   +   54/18.21  +  .........+   54/87.90

A = 54/3 . ( 1/15.18 + 1/18.21 + ........+ 1/87.90)

A = 54/3 . ( 1/15 - 1/18 + 1/18 -1/21 + ......+ 1/87 - 1/90)

A =54/3 . ( 1/15 -1/90)

A = 54/3 . 1/18 = 1

vậy A = 1

Mình nói lí thuyết cho nghe:

 Với phân số \(\frac{a-b}{a.b}\)\(\left(VD:\frac{1}{1.2};\frac{1}{2.3};\frac{1}{2015.2016};\frac{3}{15.18};\frac{3}{18.21};\frac{1}{10.11};\frac{1}{11.12};...\right)\)thì:

 \(\frac{b-a}{a.b}=\frac{b}{a.b}-\frac{a}{a.b}=\frac{1}{a}-\frac{1}{b}\left(VD:\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{3}{15.18}=\frac{1}{15}-\frac{1}{18}\right)\)

ÁP dụng để tính:

 c) \(\Rightarrow\frac{1}{4}C=\frac{1}{4}\left(\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\right)=\frac{3}{15.18}+\frac{3}{18.21}+....+\frac{3}{87.90}\)

\(\Rightarrow\frac{1}{4}C=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}=\frac{1}{15}-\frac{1}{90}\)

=> \(C=\left(\frac{1}{15}-\frac{1}{90}\right).4\)

a,\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(A=1-\frac{1}{2016}\)suy ra \(A=\frac{2015}{2016}\)

b, \(B=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)

\(B=5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(B=5\left(\frac{1}{10}-\frac{1}{70}\right)\)suy ra \(B=5.\frac{3}{35}\)

\(B=\frac{3}{7}\)

c,\(C=4.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)suy ra \(C=4.\frac{1}{18}\)

\(C=\frac{2}{9}\)

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)

\(A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

\(A=1.\left(1-\frac{1}{2016}\right)\)

\(A=1.\frac{2015}{2016}\)

\(A=\frac{2015}{2016}\)

b) \(B=\frac{5}{10.11}+\frac{5}{11.12}+...+\frac{5}{69.70}\)

\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(B=5.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(B=5.\frac{3}{35}\)

\(B=\frac{3}{7}\)

c) \(C=\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\)

\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(C=4.\frac{1}{18}\)

\(C=\frac{2}{9}\)

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

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

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

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

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

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

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

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

\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)

\(A=\frac{4}{4}\left(\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\right)\)

\(A=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)

\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)

\(A=\frac{15}{4}.\frac{1}{225}=\frac{1}{60}\)

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

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

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

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

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

\(B=2.\frac{1}{18}=\frac{1}{9}\)

Trả lời:

\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)

\(A=\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)

\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)

\(A=\frac{15}{4}.\frac{1}{225}\)

\(A=\frac{1}{60}\)

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

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

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

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

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

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

Lm phần A, còn b tự suy:

\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)

\(A=\frac{15}{4}.\frac{1}{225}=\frac{1}{60}\)

\(A=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{21}+\frac{9}{53} \)

\(A=\left(\frac{16}{7}+\frac{10}{21}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)+\frac{21}{31}\)

\(A=\frac{58}{21}+1+\frac{21}{31}\)

\(A=\frac{100}{21}\)

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

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

\(B=6\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(B=6.\frac{1}{18}\)

\(B=\frac{1}{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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