Question

\(^{x^{21}:\left(x^2+1\right)}\)Tìm số dư trong phép chia

Answer 1

dư x đó bạn a 

Answer 2

\(x^{21}=x^{21}-x+x=x\left(x^{20}-1\right)+x\)

Ta có tính chất \(a^n-b^n⋮\left(a-b\right)\left(a;b;n\inℕ^∗\right)\)

Do đó: \(x^{20}-1=\left(x^4\right)^5-1^5⋮\left(x^4-1\right)\)

Mà \(x^4-1=\left(x^2+1\right)\left(x^2-1\right)⋮\left(x^2+1\right)\)

\(\Rightarrow x^{20}-1⋮\left(x^2+1\right)\Rightarrow x\left(x^{20}-1\right)⋮\left(x^2+1\right)\)

Vậy x²¹ chia x² + 1 dư x

Answer 3

\(x^{21}=x^{19}\left(x^2+1\right)-x^{17}\left(x^2+1\right)+x^{15}\left(x^2+1\right)-x^{13}\left(x^2+1\right)+x^{11}\left(x^2+1\right)-x^9\left(x^2+1\right)+x^7\left(x^2+1\right)-x^5\left(x^2+1\right)+x^3\left(x^2+1\right)-x\left(x^2+1\right)+x\)

\(=\left(x^2+1\right)\left(x^{19}-x^{17}+x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x\right)+x\) 

 Vậy Số dư của phép chia x²¹ cho (x² + 1)  là x