$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$

$=\dfrac{3^{32}-1}{2}$

A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

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

2A = 3A - A = \(1-\frac{1}{3^8}\)

=> A = \(\frac{1-\frac{1}{3^8}}{2}\)

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

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^8}\)

\(\Rightarrow A=\left(1-\frac{1}{3^8}\right)\div2\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^8\times2}\)

A=1/3+1/3^2+1/3^3+...+1/3^8

3A=1+1/3+1/3^2+...+1/3^7

3A-A=1-1/3+1/3-1/3^2+1/3^2-1/3^3+...+1/3^7-1/3^8

2A=1-1/3^8

2A=6560/6561

Suy ra A=3280/6561

nho k cho minh voi nhe

1)     \(\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{15}\right)\)

\(=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\frac{4}{5}-\frac{3}{17}+\frac{1}{15}\)

\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{3}{17}-\frac{3}{17}\right)-\left(\frac{4}{3}-\frac{1}{15}\right)\)

\(=1+0-\frac{19}{15}\)

\(=-\frac{14}{15}\)

MÌNH

NHẦM

CHÚT

NHA

KẾT QUẢ CUỐI CÙNG LÀ:

\(-\frac{4}{15}\)

sorry!!!

còn câu 2

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

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

=>3A-A=\(\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

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

=>A=\(\frac{3^8-1}{3^8}:2=\frac{3^8-1}{2.3^8}\)

 A=1/3+1/3²+1/3³+...+1/3⁸

=>3A=1+1/3+1/3²+...+1/3⁷

=>3A-A=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

=>2A=1+1/3+1/3²+...+1/3⁷-1/3-1/3²-1/3³-...-1/3⁸

=1-1/3⁸

=\(\frac{6550}{6561}\)

=>A=\(\frac{6560}{6561}:2=\frac{6560}{6561}.\frac{1}{2}=\frac{3280}{6561}\)

A= 1/3 + 1/3^2 + ... + 1/3^8

3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)

3A=1+ 1/3 + 1/3^2+ ... +1/3^7

=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)

=> 2A= 1 - 1/ 3^8

2A= 6560/6561

A= 6560/6561 : 2

A= 3280/6561