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\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-\dfrac{1}{243}-\dfrac{1}{729}\\ \Rightarrow2A=1-\dfrac{1}{729}\\ \Rightarrow2A=\dfrac{728}{729}\\ \Rightarrow A=\dfrac{364}{729}\)
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\(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ G=\dfrac{81}{243}+\dfrac{27}{243}+\dfrac{9}{243}+\dfrac{3}{243}+\dfrac{1}{243}\\ G=\dfrac{121}{243}\)
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1/3+1/9+1/27+1/81+1/243=4/9+4/81+1/243=40/81+1/243=121/243
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Giải
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81
Vậy:
S x 3 - S = 1 - 1/81
2 S = 80/81
S = 80/81 : 2
S = 40/81
Giải
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81
Vậy:
S x 3 - S = 1 - 1/81
2 S = 80/81
S = 80/81 : 2
S = 40/81
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\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{3}{729}\)
\(=\frac{243+81+27+3}{729}=\frac{354}{729}\)
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\(\frac{1}{3}\)+ \(\frac{1}{9}\)+ \(\frac{1}{27}\)+ \(\frac{1}{81}\)+ \(\frac{1}{243}\)+ \(\frac{1}{729}\)
= \(\frac{243}{729}\)+ \(\frac{81}{729}\)+ \(\frac{27}{729}\)+ \(\frac{9}{729}\)+ \(\frac{3}{729}\)+ \(\frac{1}{729}\)
= \(\frac{\left(243+27\right)+\left(81+9\right)+\left(3+1\right)}{729}\)
= \(\frac{270+90+4}{729}\)
=\(\frac{364}{729}\)
:)
`@` `\text {Ans}`
`\downarrow`
\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}?\)
Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
`3A=`\(3\times\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
`3A =`\(3+\dfrac{3}{3}+\dfrac{3}{9}+\dfrac{3}{27}+\dfrac{3}{81}+\dfrac{3}{243}+\dfrac{3}{729}\)
`3A =`\(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
`3A - A=`\(\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
`2A =`\(3-\dfrac{1}{729}\)
`2A=`\(\dfrac{2186}{729}\)
\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)
chiu thua