Yêu cầu đề là gì vậy bạn?

Sửa đề: \(\left(1-\frac14\right)\left(1-\frac19\right)\left(1-\frac{1}{16}\right)\cdot\ldots\cdot\left(1-\frac{1}{1600}\right)\)

Ta có: \(\left(1-\frac14\right)\left(1-\frac19\right)\left(1-\frac{1}{16}\right)\cdot\ldots\cdot\left(1-\frac{1}{1600}\right)\)

\(=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\ldots\cdot\left(1-\frac{1}{40^2}\right)\)

\(=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{40}\right)\left(1+\frac12\right)\left(1+\frac13\right)\cdot\ldots\cdot\left(1+\frac{1}{40}\right)\)

\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{39}{40}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{41}{40}\)

\(=\frac{1}{40}\cdot\frac{41}{2}=\frac{41}{80}\)

`= 3/4 . 8/9 . 15/16. ... . 143/144`

`= (1.2.3...12)/(2.3.4....12) . (3.4.....12.13)/(2.3....11.12)`

`= 1/12 . 13/2 = 13/24`

=(1-1/2)(1+1/2)(1-1/3)(1+1/3)*...*(1-1/12)(1+1/12)

=1/2*2/3*...*11/12*3/2*4/3*...*13/12

=1/12*13/2=13/24

Ta có: \(1-\frac{1}{4}=\frac{3}{4}=\frac{1}{2}.\frac{3}{2}\); \(1-\frac{1}{9}=\frac{8}{9}=\frac{2}{3}.\frac{4}{3}\); \(1-\frac{1}{16}=\frac{15}{16}=\frac{3}{4}.\frac{5}{4}\);

...; \(1-\frac{1}{10000}=\frac{9999}{10000}=\frac{99}{100}.\frac{101}{100}\)

=> \(A=\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}.\frac{3}{4}.\frac{5}{4}....\frac{99}{100}.\frac{101}{100}\). Nhận thấy; Tích của 2 số liền kề thì bằng 1

=> \(A=\frac{1}{2}.\frac{101}{100}=\frac{101}{200}\)

Đáp số: \(A=\frac{101}{200}\)

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)

mk cảm ơn ah

TA có: \(A=\left(1-\frac14\right)\left(1-\frac19\right)\cdot\ldots\cdot\left(1-\frac{1}{400}\right)\)

\(=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{20}\right)\left(1+\frac12\right)\cdot\left(1+\frac13\right)\cdot\ldots\cdot\left(1+\frac{1}{20}\right)\)

\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{19}{20}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{21}{2}=\frac{1}{20}\cdot\frac{21}{2}=\frac{21}{40}\)