A=1/3+1/9+1/27+1/81+1/243
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DT
Dang Tung
CTVHS
23 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)
19 tháng 12 2020
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
Chúc bạn học tốt :))
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
17 tháng 1 2017
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(A=\frac{4}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(A=\frac{12}{9}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
Từ chỗ này làm dễ hơn rồi bạn tự làm tiếp đi nhé
31 tháng 8 2019
Ta có:\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
Xét\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Leftrightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{729}\)
\(\Leftrightarrow\frac{2}{3}A=\frac{243-1}{729}\Leftrightarrow A=\frac{3}{2}\times\frac{242}{729}=\frac{121}{243}\)
31 tháng 8 2019
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
NM
16 tháng 8 2023
\(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ G=\dfrac{81}{243}+\dfrac{27}{243}+\dfrac{9}{243}+\dfrac{3}{243}+\dfrac{1}{243}\\ G=\dfrac{121}{243}\)
PH
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 6 2023
(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))
(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))
Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)
Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2
Dãy số trên có số số hạng là: (25 - 1): 2 + 1 = 13
Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)
A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))
A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)( \(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))
A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)
A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)
Đặt B = \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\)
B\(\times\)3 =1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)
B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)
2B = \(\dfrac{242}{243}\)
B = \(\dfrac{242}{243}\): 2
B = \(\dfrac{121}{243}\)
11a + B = 11a + \(\dfrac{121}{243}\) (2)
Từ (1) và(2) ta có:
a\(\times\)13 + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)
a \(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\)
\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)
a \(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)
a \(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)
a \(\times\) 2 = \(\dfrac{109}{6075}\)
a = \(\dfrac{109}{6075}\): 2
a = \(\dfrac{109}{12150}\)
KS
5 tháng 3 2022
\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\)+\(\dfrac{1}{243}+\dfrac{1}{729}\)=\(\dfrac{1093}{729}\)
DT
5 tháng 3 2022
\(=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
\(\dfrac{729}{729}+\dfrac{243}{729}+\dfrac{81}{729}+\dfrac{27}{729}+\dfrac{9}{729}+\dfrac{3}{729}+\dfrac{1}{729}\)
\(=\dfrac{\left(729+243+81+27+9+3+1\right)}{729}=\dfrac{1084}{729}\)
23 tháng 1 2022
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\\ \Rightarrow A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\\ \Rightarrow\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}\\ \Rightarrow\dfrac{1}{3}A-A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}-1-\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^6}\\ \Rightarrow-\dfrac{2}{3}A=\dfrac{1}{3^7}-1\\ \Rightarrow A=\left(\dfrac{1}{2187}-1\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\left(-\dfrac{2186}{2187}\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\dfrac{1093}{729}\)
LK
28 tháng 2 2022
Các bạn làm như vậy với các cháu học sinh lớp 4, 5 là ko làm đc. KQ tính bằng 1093/729 là đúng nhưng PP làm chưa đúng.
Mình hướng dẫn con mình làm như thế này là phù hợp với kiến thức lớp 4:
Ta tách phân số như sau:
= (5/3-2/3) + (2/3-1/3) + (1/3-2/9) + (2/9-5/27) + (5/27-14/81) + (14/81-41/243) + (41/243-122/729)
Sau khi rút gọn ta còn:
= 5/3 - 122/729
= (5*243-122)/729
= 1093/729
#)Giải :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3A=3-1-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)
\(3A-A=\left(3-1-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(3A-A=3-1-1+\frac{1}{243}=\frac{244}{243}\)
\(A=\frac{244}{243}:\left(3-1\right)=\frac{122}{243}\)
\(A=\frac{122}{243}\)
#~Will~be~Pens~#