cho đa thức f(x)=x^99 – 3000x^98 + 3000x^97 – 3000x^96 +…. – 3000x^2 + 3000x – 1 tính f(2009)? 17/09/2021 Bởi Delilah cho đa thức f(x)=x^99 – 3000x^98 + 3000x^97 – 3000x^96 +…. – 3000x^2 + 3000x – 1 tính f(2009)?
Giải thích các bước giải: Ta có: $f(x)=x^{99}-3000x^{98}+3000x^{97}-3000x^{96}+…-3000x^2+3000x-1$ $\to xf(x)=x^{100}-3000x^{99}+3000x^{98}-3000x^{97}+…-3000x^3+3000x^2-x$ $\to f(x)+xf(x)=x^{100}-2999x^{99}+2999x-1$ $\to (1+x)f(x)=x^{100}-2999x^{99}+2999x-1$ $\to f(x)=\dfrac{x^{100}-2999x^{99}+2999x-1}{x+1}$ $\to f(2009)=\dfrac{2009^{100}-2999\cdot 2009^{99}+2999\cdot 2009-1}{2009+1}$ $\to f(2009)=\dfrac{2009^{100}-2999\cdot 2009^{99}+2999\cdot 2009-1}{2010}$ Bình luận
Giải thích các bước giải:
Ta có:
$f(x)=x^{99}-3000x^{98}+3000x^{97}-3000x^{96}+…-3000x^2+3000x-1$
$\to xf(x)=x^{100}-3000x^{99}+3000x^{98}-3000x^{97}+…-3000x^3+3000x^2-x$
$\to f(x)+xf(x)=x^{100}-2999x^{99}+2999x-1$
$\to (1+x)f(x)=x^{100}-2999x^{99}+2999x-1$
$\to f(x)=\dfrac{x^{100}-2999x^{99}+2999x-1}{x+1}$
$\to f(2009)=\dfrac{2009^{100}-2999\cdot 2009^{99}+2999\cdot 2009-1}{2009+1}$
$\to f(2009)=\dfrac{2009^{100}-2999\cdot 2009^{99}+2999\cdot 2009-1}{2010}$