a) ta có : \(N=-21x^{99}-21x^{98}-...-21x^2-21x\)
\(\Rightarrow xN=-21x^{100}-21x^{99}-...-21x^2-21x^2\)
\(\Rightarrow xN-N=-21x^{100}+21x\)
\(\Leftrightarrow\left(x-1\right)N=-21x^{100}+21x\Leftrightarrow N=\dfrac{21x-21x^{100}}{x-1}\)
\(\Rightarrow A=x^{100}-21x^{99}-21x^{98}-...-21x^2-21x+2010\)
\(=x^{100}+\dfrac{21x-21x^{100}}{x-1}+2010\)
\(=\dfrac{21x-21x^{100}+x^{101}-x^{100}+2010x-2010}{x-1}\)
\(=\dfrac{x^{101}-22x^{100}+2031x-2010}{x-1}\)
thay \(x=22\) ta có : \(A=\dfrac{22^{101}-22.22^{100}+2031.22-2010}{22-1}\)
\(=\dfrac{22^{101}-22^{101}+2031.22-2010}{21}=\dfrac{2031.22-2010}{21}=2032\)
vậy ............................................................................................................
câu b lm tương tự .
