Question

Giúp mình bài 3 với

Answer 1

h: Ta có: \(x^3+3x^2+4x+12=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

Answer 2

Answer 3

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

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

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

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

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