Tính A=1/3+1/9+1/27+1/81
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)
\(3A=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
Lấy \(3A-A=1-\dfrac{1}{729}\)
\(\Rightarrow2A=\dfrac{728}{729}\Rightarrow A=\dfrac{364}{729}\)
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
Câu trả lời hay nhất: 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
NHỚ TK MK NHA
A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049
3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
- ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
= 1 - 1/59049
2 x A = 59048/59049
A = 59048/59049 : 2
A = 29524/59049
A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049
3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683 - ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
= 1 - 1/59049
2 x A = 59048/59049
A = 59048/59049 : 2
A = 29524/59049
Ta có:\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
Xét\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Leftrightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{729}\)
\(\Leftrightarrow\frac{2}{3}A=\frac{243-1}{729}\Leftrightarrow A=\frac{3}{2}\times\frac{242}{729}=\frac{121}{243}\)
Phải là : A=1/3+1/9+1/27+1/81+1/243 ta có: 3A=1+1/3+1/9+1/27+1/81 3A-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243)=1-1/243 2A=242/243 A=242/243:2=121/243
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
\(3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
\(3A-A=\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)
\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)
A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
3A = \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
3A - A = ( \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )
2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)
A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)
A= 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Ax3=(1/3 + 1/9 + 1/27 + 1/81 + 1/243)x3
Ax3=1/3 x 3 + 1/9 x 3 + 1/27 x 3 + 1/81 x 3 + 1/243 x 3
Ax3=1+1/3+1/9+1/27+1/81
Ax3-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243)
Ax(3-1)=1-1/243
Ax2=243/243-1/243
Ax2= 242/243
A = 242/243:2
A = 242/243 x 1/2
A = 121/243
Vậy A= 121/243
Li-ke cho mik nhé!
\(3S=241+81+27+9+...+\dfrac{1}{9}+\dfrac{1}{27}\)
\(2S=3S-S=241-\dfrac{1}{81}=\dfrac{241x81-1}{81}\)
\(\Rightarrow S=\dfrac{241x81-1}{2x81}\)
Ta có: 1/3 = 27/81
1/9 = 9/81
1/27 = 3/81
Thay các giá trị vừa quy đồng vào A ta có:
A = 27/81 + 9/81 + 3/81 + 1/81
A = 27 + 9 + 3 + 1
81
A = 40
81