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\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
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`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2013.2015}=\frac{1}{2}-\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)=\frac{1}{2}-\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{2015}\right)=\frac{1}{2}-\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2014}{2015}=\frac{1}{2}-\frac{1}{x}\)
\(\Leftrightarrow\frac{1007}{2015}=\frac{1}{2}-\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1007}{2015}\Leftrightarrow\frac{1}{x}=\frac{1}{4030}\)
\(\Rightarrow x=4030\)
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a) 1/5 x 19 x 2/3 x 10 x 1/2 x 3 x 1/38
= ( 1/5 x 10 ) x ( 19 x 1/38 ) x ( 2/3 x 1/2 ) x 3
= ( 2 x 1/2 ) x ( 1/3 x 3 )
= 1 x 1
= 1
b) 5/8 x 4 x 8/5 x 7 x 0,25 x 1/8
= ( 5/8 x 8/5 ) x (4 x 0,25 ) x ( 7 x 1/8 )
= 1 x 1 x 7/8
= 7/8
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Lời giải:
$A=(99-97)+(95-93)+(91-89)+...+(7-5)+(8-1)$
$=\underbrace{2+2+2+...+2}_{24}+7$
$=2\times 24+7$
$=55$
Ta có: 99–97 +95–93+...+7–5+3−1
=(99−97)+(95−93)+...+(7−5)+(3−1)
=2+2+...+2+2 (có 25 số 2)
=2.25
=50
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a) 20,8 x 45 + 0,37 x 15 + 20,8 x 55 x 0,63
= 20,8 x ( 45 + 55 x 0,63 ) + 0,37 x 15
= 20,8 x ( 45 + 34,65 ) + 5,55
= 20,8 x 79,65 + 5,55
= 1656,72 + 5,55
= 1662,27
b) ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x ( 1 + 1/3 - 1 và 1/3 )
= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x [( 1 - 1 ) + ( 1/3 - 1/3 ) ]
= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x 0
= 0
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(1-2+3-4+.....-98+99)x(2013x6-2013-2013x5)
=(1-2+3-4+....-98+99)x[ 2013x(6-1-5)]
= (1-2+3-4+....-98+99)x(2013x0)
= (1-2+3-4+....-98+99)x0 = 0
A x 2 = 2/1 x3 + 2/ 3 x 5 + 2/ 5 x 7 + ................. + 2/ 2013 x 2015
= 1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + .................. + 1/2013 – 1/2015
= 1 – 1/2015 = 2014/2015
Vậy A = 2014/2015 : 2 = 2014/4030.
\(\frac{2014}{4030}\)