\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2022}\)

=>x+1=2022

hay x=2021

Bài làm của bạn đây nhé

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\\ =1-\dfrac{1}{2019}\\ =\dfrac{2019-1}{2019}=\dfrac{2018}{2019}\)

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.`

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

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

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

:)))))

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

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

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

\(\Leftrightarrow x=2\)

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

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

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

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

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

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

⇔ \(\dfrac{x}{1007}=\dfrac{28}{14}\)

⇔ \(\dfrac{x}{1007}=2\)

⇔ \(x=2.1007\)

⇔ \(x=2014\)

Vậy \(x=2014\)

\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)

\(\Rightarrow x+1=2017\)

\(\Rightarrow x=2017-1=2016\)

Vậy x = 2016

\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{2016}{2017}\)

1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)

\(\dfrac{3}{4}\)+\(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)

\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{2013}{8068}\)

Bn tự lm tiếp nhé!!! Sorry mk đang vội