\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+...+\frac{x}{39.41}=\frac{1}{41}\)
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LD
6
13 tháng 8 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
S
13 tháng 8 2019
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
8 tháng 4 2016
=1/2*(1-1/3+1/3-1/5+....+1/x+1/x+2)
=1/2*(1-1/x+2)
=>1/2*x+1/x+2=20/21
Đến đó đưa về giống tìm x nha
TT
tìm x : \(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
1
S
25 tháng 2 2017
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{999}=\frac{1}{2}.\frac{98}{99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{49}{99}\)
\(\frac{1}{x}=\frac{49}{99}+\frac{1}{9999}\)
\(\frac{1}{x}=\frac{50}{101}\)
\(x=1:\frac{50}{101}\)
\(x=\frac{101}{50}\)
Vậy \(x=\frac{101}{50}\)
17 tháng 3 2018
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)
\(\frac{x}{14}=\frac{16}{21}\)
\(\Rightarrow x\cdot=21=14\cdot16\)
\(\Rightarrow x\cdot21=224\)
\(\Rightarrow x=\frac{224}{21}\)
9 tháng 5 2019
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x(x+2)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x(x+2)}\right]=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right]=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\left[1-\frac{1}{x+2}\right]=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\Leftrightarrow x+2=41\Leftrightarrow x=39\)
9 tháng 5 2019
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}.\)
\(1-\frac{1}{x+2}=\frac{20}{41}\Rightarrow\frac{1}{x+2}=\frac{21}{41}=\frac{21}{21x+42}\Rightarrow21x+42=41\Rightarrow x=-\frac{1}{21}\)
29 tháng 12 2016
A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)
\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)
\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)
X=16
NN
2
BI
25 tháng 3 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{5}{11}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\Rightarrow1-\frac{1}{x+2}=\frac{5}{11}\div\frac{1}{2}=\frac{10}{11}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{10}{11}=\frac{1}{11}\Rightarrow x+2=11\Rightarrow x=11-2=9\)
25 tháng 3 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x+\left(x+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+2}\)
\(=1-\frac{1}{x+2}=\frac{5}{11}\)
\(\frac{1}{x+2}=1-\frac{5}{11}=\frac{6}{11}\)
=> không có kết quả
CB
2
TN
27 tháng 3 2016
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}=\frac{1}{x}\)
\(\Rightarrow1-\frac{1}{2005}=\frac{1}{x}\)
\(\Rightarrow\frac{2004}{2005}=\frac{1}{x}\)
tới đây tự làm nhé
\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+........+\frac{x}{39\cdot41}=\frac{1}{41}\)
\(\Rightarrow x\cdot\left[\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.........+\frac{1}{39\cdot41}\right)\right]=\frac{1}{41}\)
\(\Rightarrow x\cdot\left[\frac{1}{2}\cdot\left(1-\frac{1}{41}\right)\right]=\frac{1}{41}\)
\(\Rightarrow x\cdot\left(\frac{1}{2}\cdot\frac{40}{41}\right)=\frac{1}{41}\)
\(\Rightarrow x\cdot\frac{20}{41}=\frac{1}{41}\)
\(\Rightarrow x=\frac{1}{41}:\frac{20}{21}\)
\(\Rightarrow x=\frac{1}{41}\cdot\frac{21}{20}\)
\(\Rightarrow x=\frac{21}{820}\)
ai k mh mh k lại
k cho mh nha