Question

[Phùng Ngọc Quốc Bảo]

CM đẳng thức

\(\dfrac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}\)=\(\dfrac{1}{x^3+1}\)

Answer 1

[Nguyễn Việt Lâm]

Xét \(x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1\)

\(=\left(x^{27}+x^{21}+x^{15}+x^9+x^3\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)

\(=x^3\left(x^{24}+x^{18}+x^{12}+x^6+1\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)

\(=\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)

Vậy ta có

\(VT=\dfrac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\dfrac{1}{x^3+1}\) (đpcm)