Chọn B

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

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

\(\Leftrightarrow22< n< 24\)

\(\Rightarrow n=23\)

\(A=10^3-\left\{-5^3.2^3-11.\left[x^2-5.2^3+\left(121-11^2\right)\right]\right\}\)

\(A=10^3-\left\{\left(-10\right)^3-11.\left[x^2-40\right]\right\}\)

\(A=10^3-\left\{\left(-10\right)^3-11x^2+440\right\}\)

\(A=10^3+10^3+11x^2-440\)

\(A=2000-440+11x^2\)

\(A=1560+11x^2\)

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