\(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{1999.2000}\) =1. Tìm x
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2
HN
17 tháng 3 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)
\(\Rightarrow x=50\)
Vậy x = 50
EG
17 tháng 3 2018
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)
\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)
4 tháng 5 2019
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2020}\)
\(\Rightarrow x+1=2020\Leftrightarrow x=2019\)
Vậy x = 2019
TN
4
7 tháng 4 2016
\(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2006.2007}=\frac{2006}{2007}\)
\(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+\frac{x}{3}-\frac{x}{4}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x}{2007}-\frac{x}{2007}=\frac{2006}{2007}\)
\(2007x-x=2006\)
\(2006x=2006\)
\(x=1\)
30 tháng 7 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)\times x}=\frac{15}{16}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}=\frac{15}{16}\)
\(1-\frac{1}{x}=\frac{15}{16}\)
\(\frac{1}{x}=\frac{1}{16}\)
\(\Rightarrow x=16\)
8 tháng 4 2016
ta có 1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=17/18
1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=17/18
1-1/x+1=17/18
1/x+1=1-17/18
1/x+1=1/18
suy ra: x+1=18
x=18-1
x=17
MH
2 tháng 2 2016
\(\text{Đề }\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\left(1-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)
=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)
=> 27x - 27 = 30x - 10
=> 27x - 30x = -10 + 27
=> -3x = 17
=> x = -17/3.
1 tháng 8 2017
Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)
\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x + 1 = 100
=> x = 99
\(\Leftrightarrow x.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1999}-\frac{1}{2000}\right)=1\)
\(\Leftrightarrow x.\left(1-\frac{1}{2000}\right)=1\Leftrightarrow x\cdot\frac{1999}{2000}=1\Leftrightarrow x=\frac{2000}{1999}\)