\(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ G=\dfrac{81}{243}+\dfrac{27}{243}+\dfrac{9}{243}+\dfrac{3}{243}+\dfrac{1}{243}\\ G=\dfrac{121}{243}\)

G= 121/243

Học tốt

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

`@` `\text {Ans}`

`\downarrow`

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

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

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

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

`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 =`\(3-\dfrac{1}{729}\)

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

\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)

chiu thua

\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{13}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{40}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{121}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{364}{243}+\dfrac{1}{720}\)
\(=\dfrac{1092}{720}=\dfrac{91}{60}\)
hoặc có thể viết dưới dạng thập phân gần bằng \(1,5167\)

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

1/3+1/9+1/27+1/81+1/243=4/9+4/81+1/243=40/81+1/243=121/243

1/3+1/9+1/27+1/81+1/243=121/243

=81/243+27/243+9/243+3/243+1/243

=121/243

Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)

\(2A=1-\frac{1}{3^5}\)

\(A=\frac{242}{243}:2\)

\(A=\frac{121}{243}\)

Ủng hộ mk nha !!! ^_^

Giải
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S =  1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 =  1 + 1/3 + 1/9 + 1/27 + 1/81  
Vậy: 
S x 3 - S = 1 - 1/81
2 S = 80/81
S = 80/81 : 2 
S = 40/81

Giải
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S =  1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 =  1 + 1/3 + 1/9 + 1/27 + 1/81  
Vậy: 
S x 3 - S = 1 - 1/81
2 S = 80/81
S = 80/81 : 2 
S = 40/81

1/3+1/9+1/27+1/81+1/243

=1x81/3x81+1x27/9x27+1x3/81x3+1/243

=81/243+27/243+3/243+1/243

=112/243