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\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2022}\)
=>x+1=2022
hay x=2021
Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\\ =1-\dfrac{1}{2019}\\ =\dfrac{2019-1}{2019}=\dfrac{2018}{2019}\)
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
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
:)))))
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
\(\Rightarrow x+1=2017\)
\(\Rightarrow x=2017-1=2016\)
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
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 \(x=\dfrac{147}{50}\)
\(\dfrac{\left(1.2+2.3+3.4+...+98.99\right).x}{26950}=12\dfrac{6}{7}:\dfrac{-3}{2}\\ \Rightarrow\left(1.2+2.3+3.4+...+98.99\right).x:26950=\dfrac{90}{7}:\dfrac{-3}{2}\\ \left(1.2+2.3+3.4+...+98.99\right).x:26950=\dfrac{-60}{7}\\ \left(1.2+2.3+3.4+...+98.99\right).x=\dfrac{-60}{7}.26950\\ \left(1.2+2.3+3.4+...+98.99\right).x=-231000\\ \left\{\left[99.98.\left(98+2\right)\right]:3\right\}.x=-231000\\ 323400x=-231000\\ x=-231000:323400\\ x=\dfrac{-5}{7}\)
Đặt A=1.2+2.3+...+98.99
=>3A=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3-1.2.3+2.3.4-...-97.98.99+98.99.100
=98.99.100
=>A=98.99.100:3=323400
=>\(\dfrac{323400x}{26950}=\dfrac{90}{7}\cdot\dfrac{2}{-3}\)
<=>12x=\(-\dfrac{60}{7}\)
<=>x=\(-\dfrac{60}{12.7}\)
<=>x=\(-\dfrac{5}{7}\)
Vậy...
`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`