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

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

\(=\frac{2187}{2187}+\frac{729}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}\)

\(=\frac{2187+729+81+27+9+3+1}{2187}\)

\(=\frac{3037}{2187}\)

Đúng 100%

toán lớp 4 chưa học bình phương -.-

\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729\cdot3}\)

\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

\(3A-A=\left(4+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

\(2A=3-\frac{1}{3^7}\)

\(A=\frac{ 1}{2}\left(3-\frac{1}{3^7}\right)\)

Gia trị biểu thức là : (3-1/3^7) / 2

Tk mk nha

\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729.3}\)

\(A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

=> \(3A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

=> \(3A-A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

<=> \(2A=3-\frac{1}{3^7}=\frac{3^8-1}{3^7}\)

=> \(A=\frac{3^8-1}{2.3^7}\)

121/81

\(=\frac{121}{81}\)

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

\(\Rightarrow3A-A=3-\dfrac{1}{81}\Rightarrow2A=\dfrac{242}{81}\Rightarrow A=\dfrac{121}{81}\)

\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\)

\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)

\(2A=\dfrac{6560}{6561}\)

\(A=\dfrac{3280}{6561}\)

1:

a: =23/27-11/17+4/27+28/17

=23/27+4/27+28/17-11/17

=1+1=2

b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)

=2/3-2/9

=6/9-2/9

=4/9

c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)

=11/5(7/3-1/3)

=11/5*2

=22/5

d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)

e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)