\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)

\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=3.\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)

\(A=\dfrac{3^2}{1\times4}+\dfrac{3^2}{4\times7}+\dfrac{3^2}{7\times10}+\dfrac{3^2}{10\times13}+\dfrac{3^2}{13\times16}...+\dfrac{3^2}{97\times100}\)

\(=3\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16} +...+\dfrac{3}{97\times100}\right)\)

\(=3\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)\(=3\times\left(1-\dfrac{1}{100}\right)\)

\(=3\times\dfrac{99}{100}\)

\(=\dfrac{297}{100}\)

\(=2\dfrac{97}{100}\)

Vậy \(A=2\dfrac{97}{100}\)

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 ~

\(A=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{6}{28}=\dfrac{2}{28}=\dfrac{1}{14}\)

`3A = 3/(4.7) + 3/(7.10) + .. + 3/(25.28)`

`3A = 1/4 - 1/7 + 1/7 - 1/10 +... + 1/25 - 1/28`

`3A = 3/14`

`A = 1/14.`

1.

E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)

E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)

E = 1 - \(\dfrac{1}{22}\)

E = \(\dfrac{21}{22}\)

2.

(x - 4)(x - 5) = 0

TH_1:

x - 4 = 0 => x = 4

TH_2:

x - 5 = 0 => x = 5

Vậy: x = 4 hoặc x = 5

\(\dfrac{x}{1.4}\)+\(\dfrac{x}{4.7}+\dfrac{x}{7.10}+\dfrac{x}{10.13}+\dfrac{x}{13.16}=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\dfrac{15}{16}=\dfrac{5}{2}\) \(x=\dfrac{5}{2}:\dfrac{15}{16}\) \(x=\dfrac{80}{30}=\dfrac{8}{3}\) DAY LA BAI LAM CUA MK NHO TICK CHO MK NHA CAM ON BAN TRUOC

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)

\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=\dfrac{1}{103}\)

\(\Rightarrow x+3=103\)

\(x=103-3\)

\(x=100\)

Vậy x = 100

\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{25.28}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{3}{14}\)
\(=\dfrac{1}{14}\)
#NoSimp

29/28

sai rồi nha lmc1

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 !!!

Ta co:\(\)

\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+.....+\dfrac{2}{73.76}\)

\(=>A=2.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+....+\dfrac{1}{73.76}\right)\)

\(=>A=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{73}-\dfrac{1}{76}\right)\)

\(=>A=2.\left(1-\dfrac{1}{76}\right)\)

\(=>A=2.\dfrac{75}{76}=\dfrac{2.75}{2.38}\)

\(=>A=\dfrac{75}{38}\)

Tick cho mk nha