Tham khảo :Câu hỏi của hoàng quỳnh dương - Toán lớp 7 - Học toán với OnlineMath

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

   \(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

   \(=\frac{1}{3}-\frac{1}{39}\)

   \(=\frac{4}{13}\)

Study well ! >_<

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{37}-\frac{1}{39}\)

\(B=\frac{1}{3}-\frac{1}{39}=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(\Rightarrow B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)

\(\Rightarrow B=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}\)

a) \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)

b) \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)

\(B=\frac{1}{4}-\frac{1}{76}\)

\(B=\frac{19}{76}-\frac{1}{76}=\frac{18}{76}=\frac{9}{38}\)

a, \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{4}{13}\)

b, \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)

\(B=\frac{1}{4}-\frac{1}{76}\)

\(B=\frac{9}{38}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{37}-\frac{2}{39}\)

\(=\frac{2}{3}-\frac{2}{39}\)

\(=\frac{8}{13}\)

Ta có:

\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)

\(\frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)

\(\frac{2}{7.9}=\frac{1}{7}-\frac{1}{9}\)

\(......................................\)

\(\frac{2}{37.39}=\frac{1}{37}-\frac{1}{39}\)

nên \(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(C=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)

A=1/1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50

A=1/1-1/50

A=49/50

Vay A=49/50

B=1/3-1/5+1/5-1/7....+1/37-1/39

B=1/3-1/39

b=36/117

B=4/13

39%.21%=18% ; 1000/125.125/1000

B=1/3.5+1/5.7+1/7.9+...+1/37.39 

=1/2(2/3.5+2/5.7+2/7.9+...+2/37.39)

=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39)

=1/2(1/3-1/39)

=1/2(13/39-1/39)

=1/2.4/13

=2/13

1/3.5+1/5.7+1/7.9+....+1/37.39

=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+....+1/37-1/39)

=1/2.(1/3-1/39)

=1/2.4/13

2/13

**** bạn

\(A=\frac{1}{2}.\left(\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{37.39}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)=\frac{1}{2}.\frac{12}{39}=\frac{6}{39}\)

Ta đặt nhân tử chung nha :

  \(A=\frac{1}{2}\left(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{37.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

\(=\frac{1}{2}.\frac{12}{39}\)

\(=\frac{6}{39}\)

=\(\frac{5-3}{3\cdot5}\)+ \(\frac{7-5}{5\cdot7}\)+ \(\frac{9-7}{7\cdot9}\)+...+ \(\frac{39-37}{37\cdot39}\)

= \(\frac{5}{3\cdot5}\)- \(\frac{3}{3\cdot5}\)+ \(\frac{7}{5\cdot7}\)- \(\frac{5}{5\cdot7}\)+ \(\frac{9}{7\cdot9}\)- \(\frac{7}{7\cdot9}\)+...+ \(\frac{39}{37\cdot39}\)- \(\frac{37}{37\cdot39}\)

= \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{9}\)+...+ \(\frac{1}{37}\)- \(\frac{1}{39}\)

= \(\frac{1}{3}\)- \(\frac{1}{39}\)

=\(\frac{4}{13}\)

✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°•.¸✫ nhân A với 2 rồi phân tích như vậy được

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{37\cdot39}\)

\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+....+\frac{2}{37\cdot39}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{37}-\frac{1}{39}\)

\(2A=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)

\(A=\frac{4}{13}:2=\frac{4}{13}\cdot\frac{1}{2}=\frac{2}{13}\)

Vậy \(A=\frac{2}{13}\)

Bài làm

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)

Vậy \(A=\frac{12}{39}\)

 A  =1/3.5+1/5.7 +1/7.9 + ... +1/37.39

2.A = 2/3.5+ 2/5.7+ ....+2/37.39

2.A = 1/3- 1/5+1/5-1/7 +....+1/37-1/39

      = 1/3- 1/39

2.A= 13/39 -1/39

2.A= 12/39

A  = 12/39: 2

A= 2/13

Vậy A= 2/13

=>2A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39

=>2A=1/3-1/39=4/13

=>A=2/13