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10 tháng 9 2017
\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)
\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2G=3-\frac{1}{3^5}\)
\(2G=3-\frac{1}{243}\)
\(2G=\frac{729}{243}-\frac{1}{243}\)
\(G=\frac{728}{243}:2\)
\(G=\frac{364}{243}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)
\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)
\(1-\frac{1}{x-1}=\frac{2014}{2015}\)
\(\frac{1}{x-1}=1-\frac{2014}{2015}\)
\(\frac{1}{x-1}=\frac{1}{2015}\)
\(\Rightarrow x-1=2015\)
\(\Rightarrow x=2016\)
7 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)
\(1-\frac{1}{x+1}=\frac{499}{500}\)
\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)
=> x + 1 = 500
=> x = 500 - 1
=> x = 499
Vậy x = 499
7 tháng 8 2016
1/1.2 + 1/2.3 + 1/3.4 +...+ 1/x.(x+1)=499/500
1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 +...+ 1/x -1/(x+1) =499/500
1-1/(x+1)=499/500
=>x/(x+1)=499/500
=>x=499
5 tháng 6 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}=\frac{15}{16}\)
5 tháng 6 2018
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
SS
2 tháng 8 2015
1.
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{2012}{2013}\)
\(A=\frac{1.2.3.4.....2012}{2.3.4.5......2013}\)
\(A=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)
\(B=\frac{2012\left(2013-2012\right)}{2012\left(2011+2\right)}\)
\(B=\frac{2013-2012}{2011+2}\)
\(B=\frac{1}{2013}\)
\(Vì:\frac{ 1}{2013}=\frac{1}{2013}\)
\(\Rightarrow\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)
\(Hay: A=B\)
10 tháng 6 2018
\(A=\frac{1\times2}{2\times2}\times\frac{2\times3}{3\times3}\times\frac{3\times4}{4\times4}\times\frac{4\times5}{5\times5}\times...\times\frac{2012\times2013}{2013\times2013}\)
\(\Rightarrow A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2012}{2013}\)
\(\Rightarrow A=\frac{1\times2\times3\times4\times...\times2012}{2\times3\times4\times5\times...\times2013}\)
\(\Rightarrow A=\frac{1}{2013}\)
\(B=\frac{2012\times2013-2012\times2012}{2012\times2011+2012\times2}\)
\(\Rightarrow B=\frac{2012\times\left(2013-2012\right)}{2012\times\left(2011+2\right)}\)
\(\Rightarrow B=\frac{2012\times1}{2012\times2013}\)
\(\Rightarrow B=\frac{1}{2013}\)
NT
6
23 tháng 8 2018
\(\frac{B}{2}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\)
\(\frac{B}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(\frac{B}{2}=\frac{100}{101}\)
\(B=\frac{200}{101}\)
CT
3
H
29 tháng 4 2019
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2017}{2018}\right)\)
\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)
L
1 tháng 5 2019
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
DP
26 tháng 6 2017
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
ML
26 tháng 6 2017
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
\(\frac{1}{1x2}x\frac{4}{2x3}x\frac{9}{3x4}x...x\frac{10000}{100x101}=\frac{1x1}{1x2}x\frac{2x2}{2x3}x\frac{3x3}{3x4}x...x\frac{100x100}{100x101}\)
=\(\frac{1x2x3x...x100}{1x2x3x...x100}x\frac{1x2x3x...x100}{2x3x4x...x101}=1x\frac{1}{101}=\frac{1}{101}\)