Question

Tính (16*8-1)/(2+1)(2*2+1)(2*4+1)(2*8+1)(2*16+1)

Giúp mk vs 😉😉

Answer 1

dưới mẫu nè: (2+1)(2^2+1)(2*4+1)(2*8+1)(2*16+1)=(2*4-1)(2*4+1)(2*8+1)(2*16+1)(*vì 2+1=2*2-1)

cứ như thế thì được: 2*32-1

Answer 2

Ta có : \(\frac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

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

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

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

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