B)A*2=(1/2+1/4+....+1/256)*2

=1+1/2+1/4+....+1/128)

A*2-A=(1+1/2+1/4+...+1/128)-(1/2+1/4+...+1/256)

=1-1/256

=255/256

a) Đặt A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\)

  \(\Rightarrow\frac{1}{3}\times A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\)

Lấy \(A-\frac{1}{3}\times A\)theo vế ta có : 

\(A-\frac{1}{3}\times A=\left(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\right)-\left(\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\right)\)

\(\Rightarrow\frac{2}{3}\times A=\frac{5}{2}-\frac{5}{486}\)

\(\Rightarrow\frac{2}{3}\times A=\frac{605}{243}\)

  \(\Rightarrow A=\frac{605}{243}:\frac{2}{3}\)

  \(\Rightarrow A=\frac{605}{162}\)

Vậy  \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}=\frac{605}{162}\)

b) Đặt B = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)

=> \(\frac{1}{2}\times B=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)

Lấy B trừ \(\frac{1}{2}\times B\)theo vế ta có : 

\(B-\frac{1}{2}\times B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...++\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{512}\right)\)

\(\Rightarrow\frac{1}{2}\times B=\frac{1}{2}-\frac{1}{512}\)

\(\Rightarrow\frac{1}{2}\times B=\frac{255}{512}\)

\(\Rightarrow B=\frac{255}{512}:\frac{1}{2}\)

\(\Rightarrow B=\frac{255}{256}\)

Vậy \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}=\frac{255}{256}\)

\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^{64}-1\right)\)

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

1. 2006/987654321 + 2007/246813579 = 2007/246813579 + 2006/987654321

=>

2.

3 - (5.3/8 + X - 7 . 5/24) : 6 . 2/3 =2

3 - (15/8 + X - 35/24) : 4 = 2

3 - (15/8 + X - 35/24) = 2 . 4

3 - (15/8 + X - 35/24) = 8

15/8 + X - 35/24 = 3 - 8

15/8 + X - 35/24 = -5

15/8 + X = -5 + 35/24

15/8 + X = -85/24

X = -85/24 - 15/8

X = -65/12

Chính xác không bạn

B=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1

Vậy B<A

a)    \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=1+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)

\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)

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

\(\Leftrightarrow\)\(A=2-\frac{1}{2^7}=\frac{255}{128}\)

b)  \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\frac{2}{7}=\frac{1}{7}\)