\(A=x^{100}-12x^{99}+12x^{98}-12x^{97}+...-12x^3+12x^2-12x+12\)
Thay x = 11 ta có:
\(A=11^{100}-12.11^{99}+12.11^{98}-...-12.11^3+12.11^2-12.11+12\)
\(=11^{100}-12\left(11^{99}-11^{98}+11^{97}-...+11^3-11^2+11\right)+12\)
Đặt \(B=11^{99}-11^{98}+...+11\)
\(\Rightarrow11B=11^{100}-11^{99}+...+11^2\)
\(\Rightarrow12B=11^{100}+11\)
\(\Rightarrow B=\dfrac{11^{100}+11}{12}\)
Từ đó, \(A=11^{100}-12.\dfrac{11^{100}+11}{12}+12\)
\(=11^{100}-11^{100}-11+12=1\)
Vậy A = 1
Ta có: \(x=11\Rightarrow x+1=12\)
Khi đó, ta được:
\(A=x^{100}-12x^{99}+12x^{98}-12x^{97}+...-12x^3+12x^2-12x+12\)
\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-\left(x+1\right)x^{97}+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+12\)
\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-x^{98}-x^{97}+...-x^4-x^3+x^3+x^2-x^2-x+12\)
\(=\left(x^{100}-x^{100}\right)-\left(x^{99}-x^{99}\right)+\left(x^{98}-x^{98}\right)-...-\left(x^3-x^3\right)+\left(x^2-x^2\right)-x+12\)
\(=0-x+12=0-11+12=-11+12=1\)
Vậy tại x=11 thì A=1
