Question

a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)           b) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

c) \(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...............+\frac{1}{1280}\)             d) \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\) 

e) \(1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)            f) \(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

g) \(3+\frac{3}{5}+\frac{3}{25}+\frac{3}{125}+\frac{3}{625}\)             

Answer 1

Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?

Answer 2

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

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

2A-A = \(1-\dfrac{1}{32}\)

A=  \(\dfrac{31}{32}\)

 

Answer 3

b)\(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) 

Dặt A=

3A= \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A-A=\(1-\dfrac{1}{792}\) 

2A= \(\dfrac{791}{792}\) 

A= \(\dfrac{791}{792}:2=\dfrac{791}{1584}\)

Answer 4

c)\(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\) 

Đặt A= \(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\)

\(2A=\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{10}+....+\dfrac{1}{640}\) 

\(2A-A=\dfrac{2}{5}-\dfrac{1}{1280}\) 

\(A=\dfrac{511}{1280}\)

d) \(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{192}\)

Đặt A= \(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{192}\)

\(2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+...+\dfrac{2}{192}\)

\(2A-A=\dfrac{4}{3}-\dfrac{2}{192}\)

\(A=\dfrac{127}{96}\)

 

Answer 5

e)\(1+\dfrac{5}{4}+\dfrac{5}{8}+\dfrac{5}{16}+\dfrac{5}{32}+\dfrac{5}{64}\)

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

     2A= 1+ \((\dfrac{5}{2}+\dfrac{5}{4}+\dfrac{5}{8}+\dfrac{5}{16}+\dfrac{5}{32})\)

    2A-A = 1+ (\(\dfrac{5}{2}-\dfrac{5}{32}\))

   A= 1 + \(\dfrac{75}{32}\)

  A= \(\dfrac{107}{32}\)

f)          \(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)

Đặt B = 

     4B= \(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)

    4B - B = \(6-\dfrac{3}{512}\) 

         3B = \(\dfrac{3069}{512}\) 

             B= \(\dfrac{3069}{512}:3=\dfrac{1023}{512}\)