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

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

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

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

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

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

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

2 điểm!?

thi hay sao?

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

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

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

S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729

S = 3 - 1/729 

S= 142/729

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

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

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

$=>A=1$

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!!!

222

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

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

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

chúc bạn học tốt ạ

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

\(=\dfrac{1}{2}-\dfrac{3.\left(-6\right)}{4.5}\)

\(=\dfrac{1}{2}-\dfrac{-18}{20}\)

\(=\dfrac{1}{2}+\dfrac{9}{10}\)

\(=\dfrac{5}{10}+\dfrac{9}{10}\)

\(=\dfrac{5+9}{10}\)

\(=\dfrac{14}{10}\)

\(=\dfrac{7}{5}\)

b) \(\dfrac{\dfrac{1^0}{9}.3^2.9^3}{729}\)

\(=\dfrac{9^{-1}.3^2.9^3}{729}\)

\(=\dfrac{9^{-1}.9.9^3}{729}\)

\(=\dfrac{9^{-1+1+3}}{729}\)

\(=\dfrac{9^3}{729}\)

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

\(=1\)

Bạn có thể tự bấm máy nhé:

A=\(\left(\dfrac{5}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right):\left(\dfrac{5}{2}-\dfrac{2}{3}+\dfrac{1}{32}\right)\)

\(\Rightarrow A=\dfrac{11}{5}:\dfrac{179}{96}=\dfrac{11}{5}.\dfrac{96}{179}=\dfrac{1056}{895}\)

Chúc bạn học tốt!

=9

=9

\(\dfrac{1}{x^2+x+1}-\dfrac{1}{x-x^2}-\dfrac{x^2+2x}{x^3-1}\)

\(=\dfrac{\left(x-1\right)x}{x\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2+x+1}{x\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x^2+2x\right)x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^2-x+x^2+x+1-x^3-2x^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{1-x^3}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-\left(x^3-1\right)}{x\left(x^3-1\right)}=\dfrac{-1}{x}\)