Question

Chứng minh đa thức : A= x⁹⁹⁹⁹+x⁸⁸⁸⁸ +x⁷⁷⁷⁷ +...x¹¹¹¹ +1

chia hết cho đa thức B= x⁹ +x⁸ +x⁷+... +x +1

Answer 1

A = \(\left(x^{9999}-x^9\right)+\left(x^{8888}-x^8\right)+...+\left(x^{1111}-x\right)+\left(x^9+x^8+....+x+1\right)\)

Ta có

\(x^{9999}-x^9=x^9\left(x^{9990}-1\right)\)

Mà \(x^{9990}-1⋮x^{10}-1\)

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

\(\Rightarrow x^{9999}-x^9⋮x^9+x^8+...+x+1\)

CMTT có

\(x^{8888}-x^8;x^{7777}-x^7;...x^{1111}-x\) đều chia hết cho

\(x^9+x^8+...+x+1\)

Mặt khác

\(x^9+x^8+x^7+...+x+1⋮x^9+x^8+x^7+..+x+1\)

\(\Rightarrow A⋮B\left(ĐPCM\right)\)

Answer 2

khó quá khó