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

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

1/ 2 + 2 = 4

2/ 4 + 4 = 8

3/ 8 + 8 = 16

4/ 16 + 16 = 32

5/ 32 + 32 =64

6/ 64 + 64 =128

7/ 128 + 128 =256

8/  256 + 256 =512

9/ 521 + 512 =1033

10/ 2048 + 2048 =4096

k mình đi

4

8

16

32

64

128

256

512

2048

4096

Làm trc cho 2 câu cuối

c) \(a^2-b^2-4a+4b\)

\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)

\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)

d) \(a^2+2ab+b^2-2a-2b+1\)

\(=\left(a+b\right)^2-2\left(a+b\right)+1\)

\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)

5/2 + 5/4 + 5/8 + 5/16 + 5/32 + 5/64 

= 5(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)

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

=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

=> 2A - A = 1 - 1/64

=> A = 1 - 1/64

Do đó : 5(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) = 5(1 - 1/64) = 5 . 63/64 = 315/64 

nhanh len gium minh duoc ko

\(2A=5+\frac{5}{2}+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}.\)

\(A=2A-A=5-\frac{5}{64}=\frac{315}{64}\)

Đặt \(A=1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)

\(=5\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

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

\(\Rightarrow2\cdot B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(\Rightarrow B=2\cdot B-B=1-\frac{1}{64}=\frac{63}{64}\)

\(\Rightarrow A=5\cdot\frac{63}{64}=\frac{315}{64}\)

1+80/64+40/60+20/64+10/64+5/64

1+2 27/64

3 27/64

= 2,421875 bạn nhé

các bạn cho mình cách giải cụ thể đc ko

219/64 nhé