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dat m = 3k + r voi 0 \(\le\)r \(\le\) 2 va n = 3t + s

=> x^m  + xⁿ + 1  = x^{3k + r} + x^{3t +s} + 1 = x^3k. x^r - x^r + x^3t . x^s - x^s + x^r + x^s +1

                                                     = x^r ( x^3t -1) + x^s ( x^3t - 1) + x^r + x^s + 1

ta thay: x^3k-1 \(⋮\)  \(\left(x^2+x+1\right)\)va \(\left(x^{3t}-1\right)⋮\left(x^2+x+1\right)\) 

vay \(\left(x^m+x^n+1\right)⋮\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^r+x^s+1\right)⋮\left(x^2+x+1\right)\)voi \(0\le r;s\le2\)

\(\Leftrightarrow r=2;x=1\Rightarrow m=3k+2;n=3t+1\)

\(r=1;s=2\Rightarrow m=3k+1;n=3t+2\)

\(\Leftrightarrow mn-2=\left(3k+2\right)\left(3t+1\right)-2=9kt+3k+6t=3\left(3kt+k+2t\right)\)

\(mn-2=\left(3k+1\right)\left(3t+2\right)-2=9kt+6k+3t=3\left(3kt+2k+t\right)\)

\(\Rightarrow\left(mn-2\right)⋮3\)

ap dung:  \(m=7;n=2;\Rightarrow mn-2=12⋮3\)

\(\Rightarrow\left(x^7+x^2+1\right)⋮\left(x^2+x+1\right)\)

\(\Rightarrow\left(x^7+x^2+1\right)⋮\left(x^2+x+1\right)=x^5+x^4+x^2+x+1\)

⇒xm+xn+1=x3k+r+x3t+s+1=x3k.xr−xr+x3t.xs−xs+xr+xs+1

                                                                       =xr(x3t−1)+xs(x3t−1)+xr+xs+1

Ta thấy: (x3k−1)chia hết (x2+x+1)và (x3t−1) chia hết (x2+x+1)

Vậy: (xm+xn+1)chia hết (x2+x+1)

⇔(xr+xs+1)chia hết (x2+x+1)với 0≤r;s≤2

⇔r=2;x=1⇒m=3k+2;n=3t+1

      r=1;s=2⇒m=3k+1;n=3t+2

⇔mn−2=(3k+2)(3t+1)−2=9kt+3k+6t=3(3kt+k+2t)

      mn−2=(3k+1)(3t+2)−2=9kt+6k+3t=3(3kt+2k+t)

⇒mn−2chia hết cho 3.

Áp dụng:m=7;n=2⇒mn−2=12chia hết cho 3

⇒(x7+x2+1) chia hết cho (x2+x+1)

các bạn CTV vô đây giải bài này vs :(

câu này chịu

bố nó biết

mình mới học lớp 5 mà

a/

\(x+6y⋮17\Rightarrow5\left(x+6y\right)=5x+30y⋮17\)

\(5x+47y=\left(5x+30y\right)+17y\)

\(5x+30y⋮17\left(cmt\right);17y⋮17\Rightarrow5x+47y⋮17\)

b/

\(3x+16y⋮5\Rightarrow2\left(3x+16y\right)=6x+32y=\left(5x+30y\right)+\left(x+2y\right)⋮5\)

Mà \(5x+30y⋮5\Rightarrow x+2y⋮5\)

\(\left(mn-2\right)⋮3\Rightarrow mn\) chia cho 3 dư 2

Đặt \(m=3k+r;n=3p+q\left(p;q;r;k\in N;r\ne q;1\le r;q\le2\right)\)

Vì m;n bình đẳng nên giả sử \(m\ge n\) \(\Rightarrow r\ge q\Rightarrow r=1;q=2\)

Ta có : \(x^m+x^n+1=x^{3k+1}+x^{3p+2}+1\) 

\(=\left(x^{3k+1}-x\right)+\left(x^{3p+2}-x^2\right)+\left(x^2+x+1\right)\)

\(=x\left(x^{3k}-1\right)+x^2\left(x^{3p}-1\right)+\left(x^2+x+1\right)\)

Ta thấy \(x\left(x^{3k}-1\right)+x^2\left(x^{3p}-1\right)⋮x^3-1⋮x^2+x+1\)

\(\Rightarrow\)\(x\left(x^{3k}-1\right)+x^2\left(x^{3p}-1\right)+\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)

Hay \(x^m+x^n+1⋮x^2+x+1\)