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NK
3
LV
12 tháng 8 2016
Ta có:
1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x.(x+2) = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2)
= 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2
= 1/2.(1 - 1/x+2)
=> 1/2.(1 - 1/x+2) = 20/41
1 - 1/x+ 2 = 20/41 : 1/2
1 - 1/x+2 = 40/41
1/x+2 = 1/41
=>x + 2 = 41
=>x = 41 - 2
=>x = 39
Vậy x = 39
Ủng hộ nha
12 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=2.\frac{20}{41}\)
=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
=> \(1-\frac{1}{x+2}=\frac{40}{41}\)
=> \(\frac{1}{x+2}=1-\frac{40}{41}\)
=> \(\frac{1}{x+2}=\frac{1}{41}\)
=> \(x+2=41\)
=> \(x=41-2=39\)
NK
3
12 tháng 8 2016
Gọi tổng trên là A
1/2A= 2/1.3+1/3.5+...+1/x.(x+2)
1/2A= 1-1/x.(x+2)
A=\(\frac{1-\frac{1}{x.\left(x+2\right)}}{2}\)
12 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39
LD
1
23 tháng 7 2015
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}\)
\(=\frac{3}{7}\)
b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
VT
11
1 tháng 5 2018
( \(\frac{1}{1x3}\)+ \(\frac{1}{3x5}\)+....+\(\frac{1}{9x11}\)) x \(y\) = \(\frac{2}{3}\)
( \(\frac{2}{1x3}\)+ \(\frac{2}{3x5}\)+...+\(\frac{2}{9x11}\)) x \(y\) = \(\frac{4}{3}\) (nhân 2 vế lên với 2)
(1 - \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)- ...+ \(\frac{1}{9}\)- \(\frac{1}{11}\)) x \(y\)= \(\frac{4}{3}\)
( 1 - \(\frac{1}{11}\)) x \(y\)=\(\frac{4}{3}\)
\(\frac{10}{11}\) x \(y\) =\(\frac{4}{3}\)
\(y\) = \(\frac{4}{3}\): \(\frac{10}{11}\)
\(y\) = \(\frac{4}{3}\)x \(\frac{11}{10}\)
\(y\) =\(\frac{22}{15}\)
4 tháng 3 2016
A=1/1x3+1/3x5+1/5x7+...+1/99x101
gấp cả 2 vế lên 2 lần ta có:
Ax2=2/1x3+2/3x5+2/5x7+...+2/99x101
Ax2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
Ax2=1-1/101
Ax2=100/101
A=100/101:2=50/101
4 tháng 3 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
Chúc bạn học tốt nha !!!
9 tháng 1 2016
\(S=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\right)\)
\(S=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}\times\frac{100}{101}=\frac{50}{101}\)
LM
9 tháng 1 2016
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.100}\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=1-\frac{1}{100}\)
\(S=\frac{99}{100}\)
10 tháng 8 2015
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:2=\frac{100}{101}\times\frac{1}{2}=\frac{50}{101}\)
21 tháng 7 2015
2A = 2/1.3 +2/3.5 + 2/5.7 + ... + 2/2007.2009 + 2/2009. 2011
2A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/ 2007 - 1/2009 + 1/2009 - 1/2011
Gian uoc het ta co: 2A = 1/1 - 1/2011
2A = 2010/2011
A = 2010/2011 X 1/2
A = 1005/2011
**** mình nha
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{50}{101}\)
suy ra: \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{50}{101}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{50}{101}\)
\(\frac{1}{1}-\frac{1}{x+2}=\frac{50}{101}:\frac{1}{2}=\frac{100}{101}\)
\(\frac{1}{x+2}=1-\frac{100}{101}=\frac{1}{101}\)
suy ra: \(x+2=101\)
suy ra: \(101-2=99\)