Question

cho f(x)=x⁹⁹+x⁸⁸+x⁷⁷+...+x¹¹+1

cho g(x)=x⁹+x⁸+...+x+1

chứng minh f(x) chia hết g(x)

Answer 1

đề sai 100%

Answer 2

Ta có:

\(f\left(x\right)=x^{99}+x^{88}+x^{77}+...+x^{11}+1\)

\(f\left(x\right)=\left(x^{99}+x^{88}+x^{77}+...+x^{11}\right)+1\)

\(f\left(x\right)=\left[\left(x^9\right)^{11}+\left(x^8\right)^{11}+\left(x^7\right)^{11}+...+x^{11}\right]+1\)

Ta thấy:

\(\left(x^9\right)^{11}\) chia hết cho \(x^9\)

\(\left(x^8\right)^{11}\) chia hết cho \(x^8\)

\(..........\)

\(x^{11}\) chia hết cho \(x\)

\(1\) chia hết cho \(1\)

\(\Rightarrow f\left(x\right)\) chia hết cho \(g\left(x\right)\) ( Đpcm )