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`= 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

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

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

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

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

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

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) x 2  – x – 12                              b) x 3  – 64.

c) m 3 n 3   –   m 2 n   +   5 mn 2   –   5          d) 16 x 4  – 1.