\(A=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ =\dfrac{1}{100}+1=\dfrac{101}{100}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(A=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

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

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

\(A=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-49}{50}\)

A=1/100−1/100+1/99−1/99+1/98−1/98+1/97−...−1/3+1/2−1/2+1

=1

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-...-\dfrac{1}{2.1}\\ =\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\dfrac{99}{100}\\ =\dfrac{-98}{100}\\ =-\dfrac{49}{100}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

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

C= \(\dfrac{1}{100}-\)(\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{98.99}\)+\(\dfrac{1}{99.100}\)

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

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

= \(\dfrac{1}{100}-\dfrac{99}{100}\)

=\(\dfrac{-98}{100}=-\dfrac{49}{50}\)

Ta có:

\(=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+......+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)

sau khi giản ước ta được như sau:

=\(\dfrac{1}{100}-1\)=\(\dfrac{-99}{100}\)

Chúc bạn học tốt nhé

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

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

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

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

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

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

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

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

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

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

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

a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=1-1+\dfrac{1}{72}\)

\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)

\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)

\(=-\left(-\dfrac{173}{1287}\right)\)

\(=\dfrac{173}{1287}\)

c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

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

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

Lời giải:

Đặt \(A=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow A+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}=\frac{1}{99.100}\)

\(\Leftrightarrow A+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}=\frac{1}{99.100}\)

\(\Leftrightarrow A+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}=\frac{1}{99.100}\)

\(\Leftrightarrow A+1-\frac{1}{98}=\frac{1}{99.100}\Rightarrow A=\frac{1}{9900}-\frac{97}{98}\)

`#3107`

`a)`

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=1-\dfrac{1}{2000}\)

\(=\dfrac{1999}{2000}\)

`b)`

\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

`c)`

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

b) Sửa đề:

 \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

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

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

\(=\dfrac{1}{3}.\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

c) \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\)

\(=0\)

\(#WendyDang\)