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18 tháng 7 2016

                    Đặt \(A=\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

                    \(A=\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)

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

               \(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\right)\)

              \(A=\frac{1}{2^2}-\frac{1}{2^8}\)

           \(A=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)

          \(\Rightarrow\frac{63}{256}.x=\frac{1}{512}=\frac{1}{2^9}\)

           \(\Rightarrow\frac{63}{2^8}.x=\frac{1}{2^9}\)

            \(\Rightarrow x=\frac{1}{2^9}:\frac{63}{2^8}=\frac{1}{2^9}.\frac{2^8}{63}=\frac{1}{2.63}=\frac{1}{126}\)

           Ủng hộ mk nha !!! ^_^

                   

28 tháng 5 2016

ko bít lm bài này ak! dễ mak

28 tháng 5 2016

bày cho câu b nha!

6 tháng 7 2019

#)Giải :

\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{512}\)

\(=\frac{255}{512}\)

Lời giải 

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{512}\)

\(=\frac{255}{512}\)

5 tháng 7 2021

Có : 

A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{256}\)

\(A=\frac{128}{256}-\frac{1}{256}=\frac{127}{256}\)

2 tháng 4 2020

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{512}-\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^9}-\frac{1}{2^{10}}\)

\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^8}-\frac{1}{2^9}\)

\(3A=1-\frac{1}{2^{10}}< 1\)

\(\Rightarrow A< \frac{1}{3}\)

DD
3 tháng 3 2021

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}-\frac{1}{256}\)

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

\(A+2A=\left(\frac{1}{2}-\frac{1}{4}+...-\frac{1}{256}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-...-\frac{1}{128}\right)\)

\(3A=1-\frac{1}{256}< 1\)

\(\Rightarrow A< \frac{1}{3}\).

8 tháng 8 2019

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)

\(< =>\frac{255}{256}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(< =>\frac{1}{1}-\frac{1}{100}\)

\(< =>\frac{99}{100}\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)

\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(< =>\frac{1}{100}\)

mk chuc ban hoc tot nhe :))

3 tháng 7 2019

1) 1/1.2 + 1/2.3 + ... + 1/6.7

= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7

= 1 - 1/7

= 6/7

2) 1/2 + 1/6 + 1/12 + .. + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9

= 1 - 1/9

= 8/9

3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)

\(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)

\(\frac{1.2....2019}{2.3...2020}\)

\(\frac{1}{2020}\)

4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)

       = \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)

=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)

Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)

             A  = \(\frac{1}{2}-\frac{1}{2^9}\)

8 tháng 8 2017

Mk chịu!

17 tháng 8 2017

-_- mik ko bít