Tính
1/3+1/9+1/27+1/81+1/243
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VQ
2 tháng 5 2022
tham khảo
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
A x 3 - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729
= 1 - 1/729
A x 2 = 728/729
A = 364/729
DT
Dang Tung
CTVHS
23 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)
LA
1
19 tháng 5 2022
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
A x 3 - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729
= 1 - 1/729
A x 2 = 728/729
A = 364/729
PA
2
16 tháng 10 2023
\(3A=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
Lấy \(3A-A=1-\dfrac{1}{729}\)
\(\Rightarrow2A=\dfrac{728}{729}\Rightarrow A=\dfrac{364}{729}\)
16 tháng 10 2023
A=1/3+1/9+1/27+1/81+1/243+1/729
3A=1+1/3+1/9+1/27+1/81+1/243
3A-A=(1+1/3+1/9+1/27+1/81+1/243)-(1/3+1/9+1/27+1/81+1/243+1/729)
3A-A=1-1/3+1/3-1/9+1/9-1/27+1/27-1/81+1/81-1/243+1/243-1/729)
2A=1-1/729
2A=728/729
A=728/729/2
A=364/729
TY
5
11 tháng 9 2016
đặt S=\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
=>3S= \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
=>3S-S=\(\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)
=>s=1-1/729 = 728/729
11 tháng 9 2016
1/3+1/9+1/27+1/81+1/243+1/729=(1/3+1/9+1/81)+(1/27+1/243+1/729)=37/81+37/729=333/729+37/729=370/729
NH
0
26 tháng 2 2020
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{3}{9}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{4}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{12}{27}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{13}{27}+\frac{1}{81}+\frac{1}{243}=\frac{39}{81}+\frac{1}{81}+\frac{1}{243}=\frac{40}{81}+\frac{1}{243}\)
\(=\frac{120}{243}+\frac{1}{243}=\frac{121}{243}\)
N
5
12 tháng 5 2017
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{729}{729}+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{729+243+81+27+9+3+1}{729}\)
\(=\frac{1087}{729}\)
12 tháng 5 2017
Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3A-A=3-\frac{1}{729}\)nên \(2A=3-\frac{1}{729}\)
kHI ĐÓ \(A=\left(3-\frac{1}{729}\right):2=\frac{3}{2}-\frac{1}{1458}=\frac{2197}{1458}-\frac{1}{1458}=\frac{2196}{1458}\)
A
3
27 tháng 7 2016
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)
\(2A=1-\frac{1}{729}=\frac{728}{729}\)
\(A=\frac{728}{729}:2=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)
NL
3
23 tháng 4 2016
1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
=1+ 243/729+ 81/729 + 27/729 + 9/729 + 3/729
=1093/729
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\) ta có :
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)
\(2A=1-\frac{1}{3^5}\)
\(A=\frac{1-\frac{1}{3^5}}{2}\)
Vậy \(A=\frac{1-\frac{1}{3^5}}{2}\)
Chúc bạn học tốt ~