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\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}+\frac{1}{486}\)
\(3\times A=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\)
\(3\times A-A=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}+\frac{1}{486}\right)\)
\(2\times A=\frac{3}{2}-\frac{1}{486}\)
\(A=\frac{182}{243}\)
Bài làm :
Ta có :
\(A=\frac{1}{2}\div\frac{3}{4}+\frac{1}{6}\div\frac{3}{4}=\left(\frac{1}{2}+\frac{1}{6}\right)\div\frac{3}{4}=\frac{8}{9}\)
Bài làm :
\(a,\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}\)
\(=\left(\frac{1}{2}+\frac{1}{6}\right):\frac{3}{4}\)
\(=\frac{2}{3}:\frac{3}{4}\)
\(=\frac{8}{9}\)
Bài làm
Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$
$=1-\frac{1}{21}=\frac{20}{21}$
$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$
b) 13/17 : 8/3 - 5/17 : 8/3
= ( 13/17 - 5/17 ) : 8/3
= 7/17 * 3/8
=21/136
c) 1/6 + 1/12 + 1/20 + 1/20
= (1/6 + 1/12) + (1/20 + 1/20)
= (2/12 + 1/12) + (1/20 + 1/20)
= 3/12 + 2/20
= 1/4 +2/20
= 5/20 + 2/20
= 7/20
\(b)\frac{13}{17}\div\frac{8}{3}-\frac{5}{17}\div\frac{8}{3}\)
\(=\left(\frac{13}{17}-\frac{5}{17}\right)\div\frac{8}{3}\)
\(=\frac{8}{17}\div\frac{8}{3}\)
\(=\frac{3}{17}\)
\(c)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{20}\)
\(=\left(\frac{2}{12}+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}\right)\)
\(=\frac{1}{4}+\frac{2}{20}\)
\(=\frac{5}{20}+\frac{2}{20}=\frac{7}{20}\)
1 + 2 + 3 + ... + 2010
= [(2010 - 1) : 1 + 1] . (2010 - 1) : 2
= 2010 . 2009 : 2
= 4038090 : 2
= 2019045
1 + 2 + 3 + ... + 2010
= [ ( 2010 -1 ) : 1 + 1 ] . ( 2010 -1 ) : 2
= 2010 . 2009 :2
= 4038090 : 2
= 2019045
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