2016x^2+2017x+1=2016x^2+2016x+x+1=(x+1)*(x+2016)=0 =>x=-1 hoac x=-2016

f(2016)=2016⁸ - 2017*2016⁷ +2017*2016⁶ - 2017*2016⁵ +...+2017*2016² - 2017*2016+ 2018

         =2016⁸ -( 2016⁸ + 2016) + (2016⁷+2016) - (2016⁶ + 2016)+....+2016³+2016 -( 2016² + 2016)+2018

         =2018

Thay x=2016 thì 2017=x+1 và 2018=x+2 Do đó

\(f\left(x\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-...-\left(x+1\right)x\)\(+x+2\)

           \(=x^8-x^8-x^7+x^7+x^6-...+x^2-x^2-x+x+2\)

            \(=2\)

i do not know

Ta có: 

\(A=x^8-2017x^7+2017x^6-2017x^5+...+2017x^2-2017x+25\)

\(=\left(x^8-2016x^7\right)+\left(-x^7+2016x^6\right)+...+\left(x^2-2016x\right)-x+25\)

\(=\left(x-2016\right)\left(x^7-x^6+...+x\right)-x+25\)

Thế x = 2016 vào A ta được

\(=\left(2016-2016\right)\left(2016^7-2016^6+...+2016\right)-2016+25=-2016+25=-1991\)

A=1991

Ko biet lam quen cach lam rui hihihi ^,^

\(A=x^7\left(x-2016\right)-x^6\left(x-2016\right)+x^5\left(x-2016\right)-...+x\left(x-2016\right)-\left(x-2016\right)-2016+25=-1991\)

Lời giải:

Tại $x=2016$ thì $x-2016=0$

Khi đó:
$A=x^{2016}(x-2016)-x^{2015}(x-2016)+x^{2014}(x-2016)-x^{2013}(x-2016)+.....-x(x-2016)+x-2017$

$=x^{2016}.0-x^{2015}.0+......-x.0+2016-2017=2016-2017=-1$

P(2016)= 2016⁴-2017.2016³+2017.2016²-2017.2016+2017

P(2016)=1

mk mới học lớp 5 lên ko bit

ta có: A (x) = x^4 + 2017. x^2 + 2018

mà x^4 > hoặc = 0; 2017. x^2 > hoặc = 0; 2018 >0

=> x^4 + 2017. x^2 + 2018 >0

=> A(x) không có nghiệm

Chúc bn học tốt !!!!!

Ta có : x - 1 = 2018 - 1 = 2017 

N = x⁶ - 2017x⁵ - 2017x⁴ - 2017x³ - 2017x² - 2017x - 2017

N = x⁶ - ( x - 1 ).x⁵ - ( x - 1 ).x⁴ - ( x - 1 ).x³ - ( x - 1 ).x² - ( x - 1 ).x - ( x - 1 )

N = x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x + 1

N = 1