\(1-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\\ =1-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)\\ =1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =1-\left(1-\dfrac{1}{10}\right)\\ =1-\dfrac{9}{10}\\ =\dfrac{1}{10}\\ Vì\dfrac{1}{10}< \dfrac{1}{9}\\ Vậy1-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{9.10}< \dfrac{1}{9}\)

=1-(1-1/2+1/2-1/3+...+1/9-1/10)

=1-1+1/10

=1/10<1/9

C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)

= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2

= 1/9 - 1/10 + 1/9 - 1

= 2/9 - 11/10

= -79/90

Ta có:

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

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

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

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

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

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)

\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)

\(\Rightarrow x=-\dfrac{81}{100}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 mạng

nếu thấy đúng thì tick đúng cho mk nha

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 , biết :

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

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

-1 + \(\dfrac{4}{5}\)x = \(\dfrac{9}{5}\)

\(\dfrac{4}{5}\)x = \(\dfrac{9}{5}\) + 1 ( Quy tắc chuyển vế )

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

=> 4.x = 14

=> x = 14 : 4

=> x = 3,5

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 = \(\dfrac{31.5}{10.2}\)

x = \(\dfrac{31.1}{2.2}\)

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

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

a, đặt đề bài là A

Ta có : A=( 1-1/2+1/2-1/3+...+1/9-1/10).(x-1)+1/10.x=x-9/10

= (1-1/10).(x-1)+1/10.x

= 9/10 .( x-1 )+1/10.x

=1.x-9/10

nên x= 0 hoặc 1

với -1 nữa nha

Đầu tiên thì nhắc lại cái hằng đẳng thức cho bạn nào chưa học này: (a-b)²=a²-2ab+b²<=>a²+b²=(a-b)²+2ab

\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(=\dfrac{\left(\left(1-2\right)^2+2.1.2\right)}{1.2}+\dfrac{\left(\left(2-3\right)^2+2.2.3\right)}{2.3}+...+\dfrac{\left(\left(9-10\right)^2+2.9.10\right)}{9.10}\)
\(=\dfrac{\left(\left(-1\right)^2\right)}{1.2+2}+\dfrac{\left(\left(-1\right)^2\right)}{2.3+2}+...+\dfrac{\left(\left(-1^2\right)\right)}{9.10+2}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(=1-\dfrac{1}{10}+18\)
\(=18,9=\dfrac{189}{10}.\)

~ K chắc là đúng đâu ~