3.2/1.3.2+3.2/3.5.2+3.2/5.7.2+...+3.2/49.51

3/2(2/1.3+2/3.5+2/5.7+....+2/49.51)

3/2(1-1/3+1/3-1/5+1/5-1/7+....+1/49-1/51)

3/2(1-1/51)

3/2  .    50/51

25/17

áp dụng công thức nếu có thừa số thứ 2 ở mẫu trừ đi thừa số thứ 1 bằng số trên tử thi \(\frac{1}{a}-\frac{1}{b}\) ab ở đây là 2 thừa số ở mẫu

VD;3/1.3+3/3.5+...+3/49.51(vì tất cả mẫu trừ cho nhau đều =tử)

nên = 1/1-1/3+1/3+1/5+...+1/49-1/51

      =1-1/51

      =50/51

d/s 25/17

19333333333333465667

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

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

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

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

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

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

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

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

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

A=3/(1.3) + 3/(3.5) + 3/(5.7) +.....+ 3/(49.51) 
A=3/2 . [2/(1.3) + 2/(3.5) + 2/(5.7) +.....+ 2/(49.51)] 
A=3/2 . (1/1 - 1/3 + 1/3 - 1/5 +1/5 - 1/7 +.....+ 1/49 -1/51) 
A=3/2 . (1/1 - 1/51) 
A=3/2 . 50/51 
A=25/17. 

giup minh nha 

ta có A=3/1*3+3/3*5+3/5*7+...+3/49*51

=> A=3*1/2*(2/1*3+2/3*5+..+2/49*51)

=> A=3/2*(1-1/3+1/3-1/5+..+1/49-1/51)

=> A=3/2*(1-1/51)

=> A= 3/2* 50/51

=> A= 25/17 

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

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

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

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

                  \(=\frac{2}{3}.\frac{50}{51}=\frac{20}{51}\)

              Ủng hộ mk nha !!! ^_^

25/17 mới đúng

3/1.3 + 3/3.5 + 3/5.7 + ....... + 3/49.51

= 3 x ( 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/49.51 )

= 3 x ( 1 - 1/51 )

= 3 x      50/51

=       150/151

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

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

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

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

\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

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 ~ 

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

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

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

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

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

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

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

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

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

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

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

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

\(B=33\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(A=33\)

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

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

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

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

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

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