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

\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

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

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

\(A=\frac{4}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(A=\frac{12}{9}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

Từ chỗ này làm dễ hơn rồi bạn tự làm tiếp đi nhé

1093\729

A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

Mọi người giúp mình nha!

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

B=243/729+81/729+27/727+9/727+3/727+1/727=364/727

B=243/729+81/729+27/727+9/727+3/727+1/727=364/727

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

\(=\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{3}{729}\)

\(=\frac{243+81+27+3}{729}=\frac{354}{729}\)

1/3 + 1/9 + 1/27 + 1/81 + 1/243 +1/729 = 354/729

1/729 x ( 243 + 81 + 27 + 9 + 3 + 1) = 364/729

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

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

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

= \(\frac{270+90+4}{729}\)

=\(\frac{364}{729}\)

:)

Tính nhanh mà