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

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

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

\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)

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

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 = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

Đặt 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 = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

1093/729

Úi

2A = 3 - \(\dfrac{1}{729}=\dfrac{2186}{729}\) 

A = \(\dfrac{2186}{729}:2=\dfrac{1093}{729}\)

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

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

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

$=>A=1$

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!!!

222

\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-\dfrac{1}{243}-\dfrac{1}{729}\\ \Rightarrow2A=1-\dfrac{1}{729}\\ \Rightarrow2A=\dfrac{728}{729}\\ \Rightarrow A=\dfrac{364}{729}\)

Giải:

a) \(F=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{240}\)

\(\Leftrightarrow F=\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{15.16}\)

\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{15}-\dfrac{1}{16}\)

\(\Leftrightarrow F=\dfrac{1}{2}-\dfrac{1}{16}\)

\(\Leftrightarrow F=\dfrac{7}{16}\)

Vậy ...

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

\(\Leftrightarrow G=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\)

\(\Leftrightarrow\dfrac{1}{3}G=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}+\dfrac{1}{3^6}\)

\(\Leftrightarrow\dfrac{2}{3}G=\dfrac{1}{3}-\dfrac{1}{3^6}\)

\(\Leftrightarrow G=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^6}}{\dfrac{2}{3}}\)

\(\Leftrightarrow G=\dfrac{\left(\dfrac{1}{3}-\dfrac{1}{3^6}\right)3}{2}\)

\(\Leftrightarrow G=\dfrac{1-\dfrac{1}{3^5}}{2}\)

\(\Leftrightarrow G=\dfrac{\dfrac{3^5-1}{3^5}}{2}\)

\(\Leftrightarrow G=\dfrac{3^5-1}{3^5.2}\)

Vậy ...

c) Tương tự b)

0 

0

\(3S=241+81+27+9+...+\dfrac{1}{9}+\dfrac{1}{27}\)

\(2S=3S-S=241-\dfrac{1}{81}=\dfrac{241x81-1}{81}\)

\(\Rightarrow S=\dfrac{241x81-1}{2x81}\)

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

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

=\(\dfrac{355}{729}\)

chúc bạn học tốt ạ

1 +  1/3 + 1/9 + 1/27 + 1/81

= 1 + (1/3 + 1/27) + (1/9 + 1/81)

= 1 + (9/27 + 1/27) + (9/81 + 1/81)

= 1 + 10/27 + 10/81

= 1 + 30/81 + 10/81

= 1 + 40/81

= 121/81

ảm ơn bạn nhìu !!!!

a) \(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)

\(=\dfrac{1}{-3}\left(\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{1}{-5}\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\)

\(=\dfrac{1}{-3}.\dfrac{3}{4}+\dfrac{1}{-5}.\dfrac{5}{12}\)

\(=\left(-\dfrac{1}{4}\right)+\left(-\dfrac{1}{12}\right)\)

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

b) \(A=\dfrac{81^4.3^{10}.27^5.3^{12}}{3^{18}.9^3.243^2}\)

\(=\dfrac{9^8.9^8.9^{13}.9^{10}}{9^{16}.9^3.9^3}\)

\(=\dfrac{9^{39}}{9^{22}}\)

\(=9^{17}\)

\(A=\dfrac{81^4\cdot3^{10}\cdot27^5\cdot3^{12}}{3^{18}\cdot9^3\cdot243^2}=\dfrac{3^{16}\cdot3^{10}\cdot3^{15}\cdot3^{12}}{3^{18}\cdot3^6\cdot3^{10}}=\dfrac{3^{53}}{3^{34}}=3^{19}\)

Vậy A = 3¹⁹

Ngân Hà làm đúng phần a) nhưng làm sai phần b) nên mk chỉ làm phần b) thôi