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

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

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

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

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

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

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

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

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

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

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

Trả lời cho bạn đỗ manh tiến

a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

cứu 

okee laaa

a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=179/380

b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)

c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)

=1-1/21

=20/21

d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)

\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)

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

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

bạn hãy rút gọn vế phải: x/200=1/2.2/3.3/4......98/99.99/100

  Rồi sẽ có cái phương trình:x/200=1/100

từ đó suy ra:x/200=2/200 =>x=2

:)))))

\(\dfrac{x}{200}=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)

\(\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{100}\)

\(\Leftrightarrow x=2\)

Câu b, B=\(\dfrac{5}{1\cdot2}+\dfrac{13}{2\cdot3}+\dfrac{25}{3\cdot4}+...+\dfrac{181}{9\cdot10}\)

\(=\left(\dfrac{1}{1\cdot2}+\dfrac{4}{1\cdot2}\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{12}{2\cdot3}\right)+\left(\dfrac{1}{3\cdot4}+\dfrac{24}{3\cdot4}\right)+...+\left(\dfrac{1}{9\cdot10}+\dfrac{180}{9\cdot10}\right)\)=\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)+\left(\dfrac{4}{1\cdot2}+\dfrac{12}{2\cdot3}+...+\dfrac{180}{9\cdot10}\right)\)

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

=\(\dfrac{1}{1}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-......-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\dfrac{1}{10}+\left(2\cdot9\right)\)

=\(1-\dfrac{1}{10}+18\) \(=\dfrac{9}{10}+18\)

=18.9

a, \(\dfrac{\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}}{\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}}=\dfrac{\dfrac{15}{10}-\dfrac{4}{10}+\dfrac{1}{10}}{\dfrac{18}{12}-\dfrac{8}{12}+\dfrac{1}{12}}=\dfrac{\dfrac{15-4+1}{10}}{\dfrac{18-8+1}{12}}=\dfrac{\dfrac{12}{10}}{\dfrac{11}{12}}=\dfrac{72}{55}\)