Question

Các bạn giải giúp mình bài này với:

Chứng minh đẳng thức sau:

\(\dfrac{\left[x-1\right]\left[x^2+1\right]\left[x^4+1\right]\left[x^8+1\right]}{\left[x^2-x+1\right]\left[x^4-x^3+1\right]}=\dfrac{x^{16}+1}{x^9+1}\)

Answer 1

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

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

\(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)}{x^7-x^6+x^3+x^4-x^3+1}\)

=\(\dfrac{\left(x^8-1\right)\left(x^8+1\right)}{x^7+x^4+1}\)

\(=\dfrac{x^{16}-1}{x^7+x^4+1}\)