A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

A=\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{128}-\frac{1}{256}\right)\)

A=\(1-\frac{1}{256}\)

A=\(\frac{255}{256}\)

A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

2A = 1/2 x 2 + 1/4 x 2 + 1/8 x 2 + 1/16 x 2 +1/32 x 2 + 1/64 x 1/128 + 1/256 x 2

2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128

2A - A = ( 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 ) - ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 )

A = 1 - 1/256

A = 255/256

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

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

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

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

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

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

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

a: \(A=1999\cdot2001\)

\(=\left(2000-1\right)\left(2000+1\right)\)

\(=2000^2-1=B-1\)

=>A<B

b: \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

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

=A-1

=>B<A

c: \(A=2011\cdot2013\)

\(=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)

=B-1

=>A<B

d: \(A=4\left(3^2+1\right)\left(3^4+1\right)\cdot\ldots\cdot\left(3^{64}+1\right)\)

\(=\frac12\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\ldots\cdot\left(3^{64}+1\right)\)

\(=\frac12\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot\ldots\cdot\left(3^{64}+1\right)\)

\(=\frac12\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\cdot\left(3^{32}+1\right)\cdot\left(3^{64}+1\right)\)

\(=\frac12\left(3^{16}-1\right)\left(3^{16}+1\right)\cdot\left(3^{32}+1\right)\cdot\left(3^{64}+1\right)\)

\(=\frac12\left(3^{32}-1\right)\cdot\left(3^{32}+1\right)\cdot\left(3^{64}+1\right)\)

\(=\frac12\left(3^{64}-1\right)\cdot\left(3^{64}+1\right)=\frac12\left(3^{128}-1\right)\)

=1/2B

=>A<B

câu 1 . 125

câu 2 .97

k mình nha bn

X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128

X x 127/128                                            = 127/128

                                                        X   = 127/128 : 127/128

                                                        X   = 1

2

4

8

16

32

64

128

256

2

4

8

16

64

128

256

          k mk nhé

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

C= \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

2C = \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

2C-C = \(1-\dfrac{1}{128}\)

C= \(\dfrac{127}{128}\)

C= 12+14+18+116+132+164+112821​+41​+81​+161​+321​+641​+1281​

2C = 1+12+14+18+116+132+1641+21​+41​+81​+161​+321​+641​

2C-C = 1−11281−1281​

C= 127128128127​