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30 tháng 6 2017

\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)

\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)

\(=2\left(1-\dfrac{1}{2009}\right)\)

\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)

15 tháng 6 2018

Giải:

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2009}\)

\(=\dfrac{2008}{2009}\)

c) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{4}{7.10}+...+\dfrac{3}{94.97}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(=\dfrac{1}{1}-\dfrac{1}{97}\)

\(=\dfrac{96}{97}\)

Vậy ...

Các câu sau tương tự

16 tháng 6 2018

b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2008.1009}\)

\(=\)\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2009}=\dfrac{2009}{2009}-\dfrac{1}{2009}=\dfrac{2008}{2009}\)

10 tháng 6 2017

1)Tính

a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

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

2) tìm x

\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)

\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)

\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)

\(\dfrac{4}{5}x=0\)

\(x=0:\dfrac{4}{5}\)

\(x=0\)

b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)

\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)

\(\dfrac{2}{5}x=\dfrac{31}{10}\)

\(x=\dfrac{31}{10}:\dfrac{2}{5}\)

\(x=\dfrac{31}{4}\)

10 tháng 6 2017

1. Tính:

a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

= \(\dfrac{1}{1}-\dfrac{1}{10}\)

= \(\dfrac{10}{10}-\dfrac{1}{10}\)

= \(\dfrac{9}{10}\)

b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

= \(\dfrac{1}{1}-\dfrac{1}{100}\)

= \(\dfrac{100}{100}-\dfrac{1}{100}\)

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

2. Tìm x, biết:

a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)

\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)

\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{14}{5}\)

\(x=\dfrac{14}{5}:\dfrac{4}{5}\)

\(x=\dfrac{14}{5}.\dfrac{5}{4}\)

\(x=14.\dfrac{1}{4}\)

\(x=\dfrac{14}{4}\)

Vậy \(x=\dfrac{14}{4}\)

b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)

\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)

\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)

\(\dfrac{2}{5}x=\dfrac{62}{20}\)

\(\dfrac{2}{5}x=\dfrac{31}{10}\)

\(x=\dfrac{31}{10}:\dfrac{2}{5}\)

\(x=\dfrac{31}{10}.\dfrac{5}{2}\)

\(x=\dfrac{31}{2}.\dfrac{2}{2}\)

\(x=\dfrac{31}{2}.1\)

\(x=\dfrac{31}{2}\)

Vậy \(x=\dfrac{31}{2}\)

bài này mk tự làm ko sao chép trên mạnghihi

nếu thấy đúng thì tick đúng cho mk nhavui

22 tháng 8 2017

\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)

\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)

\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)

Bấm máy nha

22 tháng 8 2017

\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)

\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)

\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)

\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)

\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)

\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)

26 tháng 3 2017

=\(\dfrac{2^2.2^2.3^2.....9^2}{1.2^2.3^2.4^2....9^2.10}\)=\(\dfrac{2^2}{10}\)=\(\dfrac{2}{5}\)

14 tháng 8 2017

a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)

\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)

\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)

\(x=\dfrac{-9198}{4400}\)

14 tháng 8 2017

a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)

\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(x+\dfrac{206}{100}=5\)

\(x=5-\dfrac{206}{100}\)

\(x=\dfrac{147}{50}\)

Vậy \(x=\dfrac{147}{50}\)

1. So sánh: a. \(\dfrac{-18}{38}\) và \(\dfrac{-32}{68}\) b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\) với 1. 2. Tìm X, biết: a. \(-\dfrac{11}{12}\)x + \(\dfrac{3}{4}\)= \(-\dfrac{1}{6}\) b. x - 43= (57-x) - 50 c. 2x-(21.3.105-105.61)= -11.26 d. \(\left|x+1\right|\)=3 e. \(\left|2x+3\right|\)=5 3. Tính: a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009} \) b. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\) c....
Đọc tiếp

1. So sánh: a. \(\dfrac{-18}{38}\)\(\dfrac{-32}{68}\)

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

2. Tìm X, biết:

a. \(-\dfrac{11}{12}\)x + \(\dfrac{3}{4}\)= \(-\dfrac{1}{6}\)

b. x - 43= (57-x) - 50

c. 2x-(21.3.105-105.61)= -11.26

d. \(\left|x+1\right|\)=3

e. \(\left|2x+3\right|\)=5

3. Tính:

a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009} \)

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

c. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

4. Chu vi của một sân hình chữ nhật là 48m. Biết chiều dài của sân bằng 140% chiều rộng. Tính diện tích của sân hình chữ nhật đó. (giải có lời giải và phép tính đầy đủ).

5. Trên cùng một nửa mặt phẳng có bờ chứa tia Ot, vẽ các tia Om và On sao cho góc tOm = 45 độ, góc tOn = 135 độ.

a. Trong 3 tia Ot, Om, On tia nào nằm giữa hai tia còn lại? Vì sao?

b. Tính số đo góc mOn. (ko cần vẽ hình)

1

Câu 2: 

a: =>-11/12x=-1/6-3/4=-2/12-9/12=-11/12

=>x=1

b: =>x-42=57-x-50=7-x

=>2x=49

hay x=49/2

d: =>x+1=3 hoặc x+1=-3

=>x=2 hoặc x=-4

e: =>2x+3=5 hoặc 2x+3=-5

=>2x=2 hoặc 2x=-8

=>x=1 hoặc x=-4

25 tháng 7 2017

\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)

\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)

\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)

\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)

Vậy.....

\(b,\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3.\)

\(=4-1\dfrac{5}{27}.\left(-\dfrac{27}{8}\right).\)

\(=4-\dfrac{32}{27}.\left(-\dfrac{27}{8}\right).\)

\(=4-\left(-4\right).\)

\(=4+4=8.\)

Vậy.....

\(c,\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}.\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}+0+0+...+0-\dfrac{1}{100}.\)

\(=\dfrac{1}{2}-\dfrac{1}{100}.\)

\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}.\)

Vậy.....

23 tháng 6 2017

a) A = \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)

A = \(\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}.\dfrac{4.4}{4.5}\)

A = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)= \(\dfrac{1}{5}\)

b) B = \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)

B = \(\dfrac{2.3.4.5}{1.2.3.4}.\dfrac{2.3.4.5}{3.4.5.6}\)= \(\dfrac{5}{3}\)

1 tháng 5 2017

\(\dfrac{\left(1.2+2.3+3.4+...+98.99\right).x}{26950}=12\dfrac{6}{7}:\dfrac{-3}{2}\\ \Rightarrow\left(1.2+2.3+3.4+...+98.99\right).x:26950=\dfrac{90}{7}:\dfrac{-3}{2}\\ \left(1.2+2.3+3.4+...+98.99\right).x:26950=\dfrac{-60}{7}\\ \left(1.2+2.3+3.4+...+98.99\right).x=\dfrac{-60}{7}.26950\\ \left(1.2+2.3+3.4+...+98.99\right).x=-231000\\ \left\{\left[99.98.\left(98+2\right)\right]:3\right\}.x=-231000\\ 323400x=-231000\\ x=-231000:323400\\ x=\dfrac{-5}{7}\)

1 tháng 5 2017

Đặt A=1.2+2.3+...+98.99

=>3A=1.2.3+2.3.(4-1)+...+98.99.(100-97)

=1.2.3-1.2.3+2.3.4-...-97.98.99+98.99.100

=98.99.100

=>A=98.99.100:3=323400

=>\(\dfrac{323400x}{26950}=\dfrac{90}{7}\cdot\dfrac{2}{-3}\)

<=>12x=\(-\dfrac{60}{7}\)

<=>x=\(-\dfrac{60}{12.7}\)

<=>x=\(-\dfrac{5}{7}\)

Vậy...