cứu với ạ

Sửa đê: A=1/5+1/10+...+1/640+1/1280

=1/5(1+1/2+...+1/128+1/256)

Đặt B=1+1/2+...+1/128+1/256

=>2B=2+1+...+1/64+1/128

=>B=2-1/256=511/256

=>A=511/1280

1/5 + 1/5  - 1/10 + 1/10 - 1/20 + 1/20 - 1/40 + ... + 1/640 - 1/1280

= 1/5 + 1/5 - 1/1280 = 511/1280

a/ Đặt 1/5= a, ta có:

1/5 + 1/10 + 1/20 + 1/40 + ... + 1/1280

= 1/a + 1/2 x a + 1/4 x a + ... + 1/256 x a
A = 1/a + 1/2 x a + 1/4 x a + ... + 1/256 x a
2 x A = 2/a + 1/a + 1/2 x a + 1/4 x a + ... + 1/128 x a
=> A = 2/a - 1/256 x a = 2/5 - 1/1280 = 511/1280

b/ 

\(\frac{121}{27}.\frac{54}{11}=\frac{11.11.27.2}{27.11}=11.2=22\)

\(\frac{100}{21}:\frac{25}{126}=\frac{100}{21}.\frac{126}{25}=\frac{25.4.21.6}{21.25}=4.6=24\)

=> \(22< n< 24\)

=> \(n=23\)

KO HIEU GIO TAY

a) \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)

\(A.2=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{640}\)

\(A.2-A=\left(\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{640}\right)-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\right)\)

\(A=\frac{2}{5}-\frac{1}{1280}=\frac{511}{1280}\)

b) \(\frac{121}{27}.\frac{54}{11}< n< \frac{100}{21}:\frac{25}{126}\)

\(22< n< 24\)

=>  n = 23

ok bạn

1/5+1/10+1/20+...+1/1280

=1/1x5+1/

B=51​+101​+201​+401​+...+12801​

�=1⋅15+12⋅15+14⋅15+18⋅15+...+1256⋅15B=1⋅51​+21​⋅51​+41​⋅51​+81​⋅51​+...+2561​⋅51​

�=15⋅(1+12+14+18+...+1256)B=51​⋅(1+21​+41​+81​+...+2561​)

Đặt �=1+12+14+18+...+1256A=1+21​+41​+81​+...+2561​

⇒2�=2+1+12+14+...+1128⇒2A=2+1+21​+41​+...+1281​

⇒2�−�=2−1256⇒2A−A=2−2561​

�=2−1256A=2−2561​

Thay A vào B

có: �=15.(2−1256)=15⋅511256=5111280B=51​.(2−2561​)=51​⋅256511​=1280511​
 

b) Ta có: \(\frac{121}{27}.\frac{54}{11}< n< \frac{100}{21}:\frac{25}{126}\)

\(\Leftrightarrow22< n< 24\)

\(\Rightarrow n=23\)