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

a) = \(\frac{127}{96}\)

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

c) Mik bỏ nha

d) = \(\frac{1023}{512}\)

e) = \(\frac{2343}{625}\)

bạn có thể trả lời rõ ra được ko

Mk chịu!

-_- mik ko bít

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

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

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

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

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

Bạn viết đề như này sao hiểu đc

a) 19 + (29 - 9*37) - (63*9 - 29*99)

= 19 + 29 - 9*37 - 63*9 + 29*99

= 19 + 29(1 + 99) - 9(37 + 63)

= 19 + 29*100 - 9*100

= 19 + 100(29 - 9)

= 19 + 100*20

= 19 + 2000 = 2019

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

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

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

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

Bài 1:

Bài 1: 1/3+1/9+1/27+1/81+1/243+1/729

Đặt:
A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân A với 3 ta có:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)