ta có :

= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )

= 59050 + 2190 + 6570 + 270 + 810

= 59050 + ( 2190 + 810 ) + 6570 + 270

= 59050 + 3000 + 6570 + 270

= 59050 + ( 3000 + 6570 ) + 270

= 59050 + 9570 + 270

= 68620 + 270

= 68890

68890

Kết quả là 68890

Nhớ trả lời cho mình

-597871

nhầm nha

viết sai rồi 

Gọi tong trên là A

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)

\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)

\(2A=1-\frac{1}{2187}\)

\(2A=\frac{2186}{2187}\)

\(A=\frac{2186}{2187}:2\)

\(A=\frac{1093}{2187}\)

Vậy tổng A = \(\frac{1093}{2187}\)

\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)

     \(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)

=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)

<=> 2y = 3- 1/2187

=> y = \(\frac{3-\frac{1}{2187}}{2}\)

\(\text{Đ}\text{ặt} A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(\Rightarrow2187A=2187+729+243+81+27+9+3+1\)

\(\Leftrightarrow2187A=3280\)

\(\Leftrightarrow A=\frac{3280}{2187}\)

Chắc chắn 100% luôn

Đặt \(V=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(\Rightarrow3V=3.\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\right)\)

\(\Rightarrow3V=1+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\right)\)

\(\Rightarrow3V=1+V-\dfrac{1}{2187}\)

\(\Rightarrow2V=1-\dfrac{1}{2187}\)

\(\Rightarrow V=\dfrac{1093}{2187}\).

1093/2187 nha bạn

A= 1/3 + 1/9 + 1/27 + 1/81 +...+1/729 + 1/2187

3A = 1 + 1/3 + 1/9 + 1/27 + 1/81 +...+1/729 

=> 3A - A = 1 - 1/2187

$A=\dfrac{2018.2017-1}{2016.2018+2017}$

$=>A={2018.2016+2018-1}{2016.2018+2017}$

$=>A={2018.2016+2017}{2016.2018+2017}$

$=>A=1$

\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)

\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)

\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)

\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)

\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)

Chúc bạn học tốt!!!

222

364/729

364/729

Cho cái cách làm

\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\)+\(\dfrac{1}{243}+\dfrac{1}{729}\)=\(\dfrac{1093}{729}\)

\(=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

\(\dfrac{729}{729}+\dfrac{243}{729}+\dfrac{81}{729}+\dfrac{27}{729}+\dfrac{9}{729}+\dfrac{3}{729}+\dfrac{1}{729}\)

\(=\dfrac{\left(729+243+81+27+9+3+1\right)}{729}=\dfrac{1084}{729}\)

\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\\ \Rightarrow A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\\ \Rightarrow\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}\\ \Rightarrow\dfrac{1}{3}A-A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}-1-\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^6}\\ \Rightarrow-\dfrac{2}{3}A=\dfrac{1}{3^7}-1\\ \Rightarrow A=\left(\dfrac{1}{2187}-1\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\left(-\dfrac{2186}{2187}\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\dfrac{1093}{729}\)

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

         A        =       1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) + \(\dfrac{1}{729}\)

    3 \(\times\) A      = 3 + 1  + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)

3 \(\times\) A   -  A  = 3 - \(\dfrac{1}{729}\)

A \(\times\)(3-1)      = \(\dfrac{2186}{729}\)

A \(\times\) 2          =    \(\dfrac{2186}{729}\)

A                 =   \(\dfrac{2186}{729}\): 2

A                  =   \(\dfrac{1093}{729}\)