Question

Chứng minh rằng f(x) chia hết cho g(x) với :

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

\(g\left(x\right)=x^9+x^8+x^7+....+x+1\)

Answer 1

Sửa lại đề bài nhé . \(f\left(x\right)=x^{99}+x^{88}+x^{77}+...+x^{11}+1\)

Xét hiệu \(f\left(x\right)-g\left(x\right)=x^9\left(x^{90}-1\right)+x^8\left(x^{80}-1\right)+x^7\left(x^{70}-1\right)+...+x\left(x^{10}-1\right)\)  

                                      \(=x^9\left[\left(x^{10}\right)^9-1\right]+x^8\left[\left(x^{10}\right)^8-1\right]+x^7\left[\left(x^{10}\right)^7-1\right]+...+x\left(x^{10}-1\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)⋮\left(x^{10}-1\right)\)

Mà \(x^{10}-1=\left(x-1\right)\left(x^9+x^8+x^7+...+x+1\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)⋮g\left(x\right)\Rightarrow f\left(x\right)⋮g\left(x\right)\)

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