tính A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
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= 1 . 1/2 + 1/2 . 1/3 + ... + 1/99 . 1/100
= 1 . 1/100
= 1/100
SAI thi mai len bao sai cho nao nha !!!!
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
= \(1-\frac{1}{2017}\)
= \(\frac{2016}{2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{2016}+\frac{1}{2016}\right)-\frac{1}{2017}\)
\(A=1+0+0+...+0-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2017}{2017}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Vậy: \(A=\frac{2016}{2017}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(A=1-\frac{1}{20}\)
\(A=\frac{19}{20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}\)
\(=\frac{19}{20}\)
a) = 1-1/2+1/2-1/3+1/3-1/4
= 1-1/4=3/4
b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018
=1-1/2018=2017/2018
c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015
= 1/2-1/2015=2015/4030-2/4030=2013/4030
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{9}-\frac{1}{10}\right)\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)
dế mà em, giải thế này nè
A=1-1/2 +1/2-1/3 +1/3-1/4 +......+1/9-1/10
A=1-1/10+9/10
Dãy số trên có tất cả số 1 là :
[ ( 96 - 1 ) : 1 + 1 ] : 2 = 48 ( số )
TA có :
( 1 + 1 + 1 + ... + 1 ) + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{95.96}\right)\)
= 48 + \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{95}-\frac{1}{96}\right)\)
= 48 + \(\left(1-\frac{1}{96}\right)\)
= 48 + \(\frac{95}{96}\)
= \(48\frac{95}{96}\)
TK nha !
Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
=> A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20
=>A= 1-1/20=19/20
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
= 1-\(\frac{1}{50}\)
= \(\frac{49}{50}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)
A=\(1-\frac{1}{4}\)
A=\(\frac{3}{4}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{6}{12}+\frac{2}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)