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

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

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

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

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}=\frac{49}{50}\)

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

= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)

= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)

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

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

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

= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\) 

= \(\frac{9}{38}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

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

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

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

\(=\dfrac{1}{3}-\dfrac{1}{39}\)

\(=\dfrac{12}{39}\)

Vậy \(A=\dfrac{12}{39}\)

b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)

\(=1-\dfrac{1}{76}\)

\(=\dfrac{75}{76}\)

Vậy \(B=\dfrac{75}{76}\)

a) Ta có :

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

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

\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)

b) Ta có :

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)

\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)

~ Học tốt ~

x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

Đề bài :

a) dãy các phân số trên có phải theo quy luật ko ?

b) tính tổng các phân số của dãy trên

1) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\)

\(=\dfrac{49}{50}\)

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

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

\(=\dfrac{1}{3}-\dfrac{1}{39}\)

\(=\dfrac{13}{39}-\dfrac{1}{39}=\dfrac{12}{39}=\dfrac{4}{13}\)

3) \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{73.76}\)

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

\(=\dfrac{1}{4}-\dfrac{1}{76}\)

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

1)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)

2)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\\ =\dfrac{1}{3}-\dfrac{1}{39}\\ =\dfrac{13}{39}-\dfrac{1}{39}\\ =\dfrac{12}{39}=\dfrac{4}{13}\)

3) \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{73.76}\\ =\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{79}\\ =\dfrac{1}{4}-\dfrac{1}{79}\\ =\dfrac{75}{316}\)

phần a thêm một câu hỏi nữa là : nếu có quy luật là gì ? .

MONG MN GIÚP ĐỠ Ạ !!!

A = 1 /1.2 + 1/ 2.3 + 1 /3.4 + . . . + 1/ 49.50 + 1/ 50.51

 A = 2 − 1/ 1.2 + 3 − 2 /2.3 + 4 − 3 /3.4 + . . . + 50 − 49 /49.50 + 51 − 50/ 50.51

A = 1 − 1/ 2 + 1/ 2 − 1 /3 + 1 /3 − 1/ 4 + . . . + 1 /50 − 1 /51

A=1-1/51

A=50/51

Cảm ơn bn

B=6/4.7+6/7.10+6/10.13+...+6/73.76

=2.(3/4.7 +3/7.10 +3/10.13 +...+3/73.76 )

=2.(1/4 −1/7 +1/7 −1/10 +1/10 −1/13 +...+1/73 −1/76 )

=2.(1/4-1/76)=2.9/38=9/19

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sau 3 phút có kết quả tùy bạn

A=\(\frac{37}{78}\)

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