Bài 4 :Cho đa thức: f(x) = x^6 – 2020.x^5 + 2020.x^4 – 2020.x^3 + 2020,x^2 – 2020.x + 2020
Tính f(2019).
Bài 4 :Cho đa thức: f(x) = x^6 – 2020.x^5 + 2020.x^4 – 2020.x^3 + 2020,x^2 – 2020.x + 2020 Tính f(2019).
By Savannah
By Savannah
Bài 4 :Cho đa thức: f(x) = x^6 – 2020.x^5 + 2020.x^4 – 2020.x^3 + 2020,x^2 – 2020.x + 2020
Tính f(2019).
Biểu thức $f(x) = x^6 – 2020x^5 + 2020x^4 – 2020x^3 + 2020x^2 – 2020$ ta có:
$x^5(x-2019)-x^4(x-2019)+x^3(x-2019)-x^2(x-2019)+x(x-2019)-(x-2019)+1$
$=(x-2019)(x^5-x^4+x^3-x^2+x-1)+1$
Thay $f(2019) $ vào biểu thức $(x-2019)(x^5-x^4+x^3-x^2+x-1)+1$ ta có:
$f(2019)=(2019-2019)(2019^5-2019^4+2019^3-2019^2+2019-1)+1$
$=0-1$
$=1$
Vậy $f(2019)=1$
Đáp án+Giải thích các bước giải:
`f(x)=x^6-2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020`
`=x^6-(2019+1).x^5+(2019+1).x^4-(2019+1).x^3+(2019+1).x^2-(2019+1).x+2019+1`
`=x^6-2019x^5-x^5+2019x^4+x^4-2019x^3-x^3+2019x^2+x^2-2019x-x+2019+1`
____
Ta có: `f(2019)=2019^6-2019.2019^5-2019^5+2019.2019^4+2019^4-2019.2019^3-2019^3+2019.2019^2+2019^2-2019.2019-2019+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`
`=1`
Vậy `f(2019)=1`