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

(4n + 5) : 3 - 121 : 11 = 4

(4n + 5) : 3 - 11 = 4

(4n + 5) : 3 = 4 + 11

4n + 5) : 3 = 15

4n + 5 = 15 × 3

4n + 5 = 45

4n = 45 - 5

4n = 40

⇒n = 10

(2n-1)² = 121

=> 2n-1 = 11

2n = 11+1

2n = 12

n = 12:2

n = 6

\(\left(2n-1\right)^2=121\)

\(\Leftrightarrow\left(2n-1\right)^2=\orbr{\begin{cases}11^2\\\left(-11\right)^2\end{cases}}\)

Do \(n\in N\)\(\Rightarrow\)\(\left(2n-1\right)^2=11^2\)

\(\Leftrightarrow2n-1=11\)

\(\Leftrightarrow2n=12\)

\(\Leftrightarrow n=6\)

(2n-1)^2=121

=>(2n-1)^2=11^2

=>2n-1=11

2n=12

n=6

Vậy n=6

121 : 11ⁿ = 1331

=> 11ⁿ = 161051

=> 11ⁿ = 11⁵

=> n = 5

\(121:11^n=1331\)

\(11^n=\frac{121}{1331}\)

\(11^n=\frac{1}{11}\)

\(11^n=11^{-1}\)

\(\Rightarrow n=-1\)

Vậy \(n=-1\)

Vậy \(n=-1\)

 n-121 = 0 

   n = 121 

sai thì đừng bầm sai nhé

mik ko chắc đâu đấy

học tốt

121.11ⁿ=1331

11ⁿ = 1331:121

11ⁿ = 11

=> n = 1

copy cái bài trên mạng ak :) có đáp án rồi mờ :) đăng lên làm j ? :))