Ta có : 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

Vậy \(A=\frac{25}{17}\)

Chúc bạn học tốt ~ 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\frac{4}{21}\)

\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)

\(B=33\)

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(C=\frac{1}{2}.\frac{98}{99}\)

\(C=\frac{49}{99}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{21}\)

\(A=1-\frac{1}{51}\)

\(A=\frac{51}{51}-\frac{1}{51}\)

\(A=\frac{50}{51}\)

\(A=\frac{21}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(A=\frac{21}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)

\(A=\frac{21}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(A=\frac{21}{4}.33.\frac{4}{21}\)

\(A=33\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\frac{98}{99}\)

\(A=\frac{49}{99}\)

\(\frac{1}{1.3}+\left|\frac{-1}{3.5}\right|+\left|\frac{-1}{5.7}\right|+...+\left|\frac{-1}{49.51}\right|=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(1-\frac{1}{51}\right)=\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)

    \(\frac{1}{1.3}+\left|-\frac{1}{3.5}\right|+\left|-\frac{1}{5.7}\right|+...+\left|-\frac{1}{49.51}\right|\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{50}{51}\)

\(=\frac{25}{51}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)

\(A=1-\frac{1}{51}\)

\(A=\frac{50}{51}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)

\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(2A=3\left(1-\frac{1}{51}\right)\)

\(2A=3.\frac{50}{51}\)

\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'

\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

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

\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{4}{8}-\frac{1}{8}\)

\(=\frac{3}{8}\)

\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)

\(=1-0-0-0-...-0-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\frac{50}{51}\)

\(=\frac{25}{17}\)

\(d,\)giống câu a tự làm nha mỏi tay quá.

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)

=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)

=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)

1/2*3+1/3*4+1/4*5+...+1/7*8

1/2-1/3+1/3-1/4+1/4-1/5-...-1/8

1/2-1/8=3/8

1/4-1/7+1/7-1/10+1/10-1/13-...-1/52                                   49/52 bạn nhé

1/4-1/52=3/13

câu này mình gọi nó là S

ta có S:2=2/1*3+2/3*5+...+2/49*51

1/1-1/3+1/3-1/5+...+1/49-1/51

1/1-1/51=50/51

S=50/51*2=100/51

1/100-1/101+1/101-1/102+1/102-1/103+...+1/2010-1/2011

1/100-1/2011

bạn tích đi nhé mình còn phải đi học bạn k cho mình nhé

theo công thức, ta tính đc: 

A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51

=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )

1-1/3=2/3 chứ ko phải 1-1/3=1/3 đâu nha bạn

theo bài ra ta có:

A=1-1/3+1/3-1/5+1/5-1/7+......+1/47-1/49+1/49-1/51

A=1-1/51

C=\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

C=\(\frac{1}{6}\left\{\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{31.37}\right\}\)=\(\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}+....+\frac{1}{31}-\frac{1}{37}\right)\)

C=\(\frac{1}{6}\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{36}{222}=\frac{6}{37}\)

D=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{49.51}\)

D=\(\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

D=\(\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

D=\(\frac{3}{2}\left(1-\frac{1}{51}\right)=\frac{3}{2}.\frac{50}{51}\)

D=\(\frac{150}{102}\)=\(\frac{25}{17}\)

a, \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b, \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

Gọi biểu thức trên là A 

\(A=\frac{17}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{17}{2}.\left(1-\frac{1}{51}\right)\)

\(A=\frac{17}{2}.\frac{50}{51}\)

\(A=\frac{25}{3}\)