## Cho đa thức : $\begin{array}{I}f(x)=x^6-2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020\end{array}$ Tính $f(2019)$

Question

Cho đa thức : $\begin{array}{I}f(x)=x^6-2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020\end{array}$
Tính $f(2019)$

in progress 0
1 năm 2021-10-27T21:25:07+00:00 2 Answers 4 views 0

1. Đáp án: $f(2019) = 1$

Giải thích các bước giải:

$f(x) = x^{6} – 2020x^{5} + 2020x^{4} – 2020x^{3} + 2020x^{2} – 2020x + 2020$

$= x^{5}(x – 2019) – x^{4}(x – 2019) + x^{3}(x – 2019) – x^{2}(x – 2019) + x(x – 2019) – (x – 2019) + 1 = (x – 2019)(x^{5} – x^{4} + x^{3} – x^{2} + x – 1) +1$

$⇒ f(2019) = (2019 – 2019)(2019^{5} – 2019^{4} + 2019^{3} – 2019^{2} + 2019 – 1) +1$

$= 0 + 1 = 1$

2. Giải thích các bước giải:

f(x)=x^6-2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020

=>f(2019)=2019^6-2020.2019^5+2020.2019^4-2020.2019^3+2020.2019^2-2020.2019+2020

=2019^6-(2019+1).2019^5+(2019+1).2019^4-(2019+1).2019^3+(2019+1).2019^2-(2019+1).2019+2019+1

=2019^6-2019^6-2019^5+2019^5+2019^4-2019^4-2019^3+2019^3+2019^2-2019^2-2019+2019+1

=(2019^6-2019^6)+(-2019^5+2019^5)+(2019^4-2019^4)+(-2019^3+2019^3)+(2019^2-2019^2)+(-2019+2019)+1

=0+0+0+0+0+0+1

=1

Vậy f(2019)=1