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Lời giải:
\(A=x^{10}+20x^9+20x^8+...+20x^3+20x^2+20x\)
\(=x^{10}+21x^9+21x^8+....+21x^3+21x^2+21x-(x^9+x^8+...+x^3+x^2+x)\)
\(=x^{10}-x.x^9-x.x^8-...-x.x^3-x.x^3-x.x-(x^9+x^8+...+x^3+x^2+x)\)
\(=-(x^9+x^8+....+x^2)-(x^9+x^8+x^3+x^2+x)\)
\(=-2(x^2+x^3+...+x^9)-x\)
\(Ax=-2(x^3+x^4+...+x^{10})-x^2\)
\(Ax-A=-2(x^3+x^4+...+x^{10})-x^2+2(x^2+...+x^9)+x\)
\(A(x-1)=x^2+x-2x^{10}\)
\(A=\frac{x^2+x-2x^{10}}{x-1}=\frac{21^2-21-2.21^{10}}{-22}=\frac{21^{10}-210}{11}\)
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Ta có: \(x=-24\Leftrightarrow-x=24\Leftrightarrow1-x=25\)
Thay vào E ta được:
\(E=x^{20}+\left(1-x\right)x^{19}+\left(1-x\right)x^{18}+...+\left(1-x\right)x^2+\left(1-x\right)x+\left(1-x\right)\)
\(E=x^{20}+x^{19}-x^{20}+x^{18}-x^{19}+...+x^2-x^3+x-x^2+1-x\)
\(E=1\)
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Sửa đề: x=21
=>x-1=20
\(A=x^{10}+x^9\left(x-1\right)+x^8\left(x-1\right)+...+x^3\left(x-1\right)+x^2\left(x-1\right)+x\left(x-1\right)\)
\(=x^{10}+x^{10}-x^9+x^9-x^8+...+x^3-x^2+x^2-x\)
\(=2x^{10}-x=2\cdot21^{10}-21\)