2

4

8

16

32

64

128

256

2

4

8

16

64

128

256

          k mk nhé

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

tk m nhé

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

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

2

4

8

16

32

64

128

256

1+1=2

2+2=44+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

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​
 

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

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

\(2A-A=1-\frac{1}{128}\)

\(A=\frac{127}{128}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{127}{128}\)

Bạn xem ở đây nha: http://olm.vn/hoi-dap/question/3155.html

\(\frac{127}{128}\)

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

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

=1/1-1/128

=127/128

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

=127/128

kq = \(\frac{127}{128}\)Bạn chỉ cần bấm máy tính là ra bài này dễ mà hihi :D :))

= 0,9921875

kick mình nha dt đó

thông cảm bài này mình chịu nha bạn

Gọi biểu thức trên là A ta có

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

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

=> \(A=2A-A\)\(=1-\frac{1}{128}\)

Vậy \(A=1-\frac{1}{128}\)