\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)

\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)

Lời giải:

$x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}+\frac{1}{100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}+\frac{1}{100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}$

$=1$

`# \text {DNamNgV}`

\(x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{98\cdot99}=\dfrac{1}{100}+\dfrac{1}{99\cdot100}\)

\(\Rightarrow x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)=\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)

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

\(\Rightarrow x-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}\)

\(\Rightarrow x-\dfrac{98}{99}=\dfrac{1}{99}\)

\(\Rightarrow x=\dfrac{1}{99}+\dfrac{98}{99}\)

\(\Rightarrow x=\dfrac{99}{99}\)

\(\Rightarrow x=1\)

Vậy, `x = 1.`

`x/(1.2)+x/(2.3)+x/(3.4)+.....+x/(2017.2018)=1`

`-> x/1 - x/2 +x/2-x/3+x/3-x/4+........+x/2017-x/2018=1`

`-> x-x/2018=1`

`-> 2017/2018 .x=1`

`-> x=2018/2017`

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

:)))))

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

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

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

\(\Leftrightarrow x=2\)

A=2022(1/1-1/2+1/2-1/3+...+1/2021-1/2022)

  =2022(1/1-1/2022)

 =2022.2021/2022

ket qua tu tinh nha

A = \(\dfrac{2022}{1.2}+\dfrac{2022}{2.3}+\dfrac{2022}{3.4}+...+\dfrac{2022}{2021.2022}\)

= \(\dfrac{2022}{1}-\dfrac{2022}{2}+\dfrac{2022}{2}-\dfrac{2022}{3}+\dfrac{2022}{3}-\dfrac{2022}{4}+...+\dfrac{2022}{2021}-\dfrac{2022}{2022}\)

= \(\dfrac{2022}{1}-\dfrac{2022}{2022}\)

= \(2021\)

Chúc bạn học tốt!! ^^

2022.( 1/1.2+ 1/2.3+...+ 1/2021+2022)

2022. ( 1-1/2+1/2-1/3+......+1/2021-1/2022)

2022.( 1-1/2022)

2022.2021/2022

2021

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

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

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

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

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

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

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

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

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

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

\(\dfrac{x}{1007}-\dfrac{1}{1.2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-...-\dfrac{1}{13.14}=\dfrac{15}{14}\)

⇔ \(\dfrac{x}{1007}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\right)=\dfrac{15}{14}\)

⇔ \(\dfrac{x}{1007}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{13}-\dfrac{1}{14}\right)=\dfrac{15}{14}\)

⇔ \(\dfrac{x}{1007}-\left(1-\dfrac{1}{14}\right)=\dfrac{15}{14}\)

⇔ \(\dfrac{x}{1007}-\dfrac{13}{14}=\dfrac{15}{14}\)

⇔ \(\dfrac{x}{1007}=\dfrac{15}{14}+\dfrac{13}{14}\)

⇔ \(\dfrac{x}{1007}=\dfrac{28}{14}\)

⇔ \(\dfrac{x}{1007}=2\)

⇔ \(x=2.1007\)

⇔ \(x=2014\)

Vậy \(x=2014\)