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1.
Ta có:
* 279
= ( 3.9)9
= 39.99
= 3.38.99
=3.(32)4.99
= 3.94.99
= 3.913
* 817
= (92)7
= 914
M = 817 - 279 - 913
= 914 + 3.913 – 913
=913(9 – 3 – 1)
= 5.912 chia hết cho 5 và 9 => chia hết cho45
Vậy M chia hết cho 45.
(y + \(\dfrac{1}{3}\)) + ( y + \(\dfrac{1}{9}\)) + ( y + \(\dfrac{1}{27}\)) + ( y + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)
( y + y + y + y ) + (\(\dfrac{1}{3}\)+ \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)
4\(y\) + ( \(\dfrac{27}{81}\) + \(\dfrac{9}{81}\) + \(\dfrac{3}{27}\) + \(\dfrac{1}{81}\) ) = \(\dfrac{56}{81}\)
4y + \(\dfrac{40}{81}\) = \(\dfrac{56}{81}\)
4y = \(\dfrac{56}{81}\) - \(\dfrac{40}{81}\)
4y = \(\dfrac{16}{81}\)
y = \(\dfrac{16}{81}\) : 4
y = \(\dfrac{4}{81}\)
\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)
\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)
\(\Rightarrow4\times y=\dfrac{16}{81}\)
\(\Rightarrow y=\dfrac{16}{81}:4\)
\(\Rightarrow y=\dfrac{4}{81}\)
M = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
=> 3M = \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)
=> 3M - M = ( \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\) ) - ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\))
2M = 1 - \(\frac{1}{6561}\)
2M = \(\frac{6560}{6561}\)
=> M = \(\frac{3280}{6561}\)
\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)
\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)
\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)
\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)
\(\Rightarrow2M=1-\frac{1}{3^8}\)
\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)
Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)
Đặt A=1/3 + 1/9 + 1/27 + 1/81 + 1/24 + 1/729
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^5}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(2A=1-\frac{1}{3^6}\)
\(A=\frac{1-\frac{1}{3^6}}{2}\)
\(\text{Đ}\text{ặt}:A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=3-\frac{1}{729}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2=\frac{1093}{729}\)
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)
1+1/3+1/9+1/27+1/81 (ko quy đồng nha)
Không quy đồng
Ta có : \(1+\frac{1}{3}-\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
ta có : \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\)
\(\Rightarrow3A-A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\)
\(\Rightarrow2A=3-\frac{1}{3^4}\)
\(\Rightarrow A=\left(3-\frac{1}{3^4}\right):2=\frac{121}{81}\)