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

chào bn

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C          =        \(\dfrac{3}{2}\) + \(\dfrac{3}{4}\) + \(\dfrac{3}{8}\)+ \(\dfrac{3}{16}\)+...........+\(\dfrac{3}{128}\)

C\(\times\)2      =  3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{4}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{16}\)+...+\(\dfrac{1}{64}\)

C\(\times\)2 - C =  3 - \(\dfrac{3}{128}\)

C           = \(\dfrac{381}{128}\)

Câu E em xem lại đề nhé 

đặt 3 ra ngoài,,,,đặt bên trong là A ,,rồi nhân A vs 1/2 ,,lấy A-1/2A=,,,,,,,  đc bao nhiu chia 1/2

   3/2 + 3/8 + 3/32 + 3/128 + 3/512

= 768/512 + 182/512 + 48/512 + 12/512 + 3/512

= 960/512 + 60/512 + 3/512

= 1023/512 

0,856m=?m

\(...=\dfrac{3}{2}x\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\left(\dfrac{256}{256}+\dfrac{64}{256}+\dfrac{16}{256}+\dfrac{4}{256}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\dfrac{341}{256}=\dfrac{1023}{512}\)

Đặt tổng trên = A

\(A=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(A.4=6+\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\)

\(A.4-A=\left(6+\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\right)-\left(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\right)\)

\(A.3=6-\frac{3}{512}=\frac{3069}{512}\)

\(A=\frac{3069}{512}:3=\frac{1023}{512}\)

Đặt A = 3/2 + 3/8 + ... + 3/512
bn tách 
3/2 = 3/2^1
3/8 = 3/2^3
...
3/512 = 3/2^9
Rồi nhân nó lên trừ được bao nhiêu - đi A ban đầu là đc
k nka

lấy 6 đâu ra bạn 

=1023/512