Lời giải:

Tại $x=2016$ thì $x-2016=0$

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

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

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

x=2016 nên x+1=2017

\(f\left(x\right)=x^{99}-x^{98}\left(x+1\right)+x^{97}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-1\)

\(=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}+...-x^3-x^2+x^2+x-1\)

=x-1=2015

\(f\left(x\right)=x^{99}-2017x^{98}+2017x^{97}-...+2017x-1\)

\(=x^{99}-2016x^{98}-x^{98}+2016x^{97}+...-x^2+2016x+x-2016+2015\)

\(=x^{98}\left(x-2016\right)-x^{97}\left(x-2016\right)+...-x\left(x-2016\right)+\left(x-2016\right)+2015\)

\(=\left(x^{98}-x^{97}+...-x+1\right)\left(x-2016\right)+2015\)

\(\Rightarrow f\left(2016\right)=2015\)

Vậy...

\(f\left(x\right)=x^{99}-2017^{x^{98}}+2017^{x^{97}}-...+2017x-1\)

\(f\left(2016\right)=2016^{99}-2017.2016^{98}+2017.2016^{97}-...+2017.2016-1\)

\(f\left(2016\right)=2016^{99}-\left(2016+1\right).2016^{98}+\left(2016+1\right).2016^{97}-...+\left(2016+1\right).2016-1\)

\(f\left(2016\right)=2016^{99}-2016^{99}-2016^{98}+2016^{98}+2016^{97}-2016^{97}-2016^{96}+...+2016^2+2016-1\)

\(f\left(2016\right)=2016-1\)

\(f\left(2016\right)=2015\)

a/ Với \(x=2016\Rightarrow2017=x+1\)

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2025\)

\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2025\)

\(A=2025-x=9\)

b/ Với \(x=-1\Rightarrow\left\{{}\begin{matrix}x^{2k}=1\\x^{2k+1}=-1\end{matrix}\right.\) ta có:

\(Q=2017-2016+2015-2014+...+3-2+1\)

\(Q=1+1+1+...+1+1\) (có \(\frac{2016}{2}+1=1009\) số 1)

\(Q=1009\)

Ta có:

\(x^6-2017x^5+2017x^4-2017x^3+2017x^2-2017x+2017\)

\(=x^6-2016x^5-x^5+2016x^4+x^4-2016x^3-x^3+2016x^2+x^2-2016x-x+2017\)

\(=x^5\left(x-2016\right)-x^4\left(x-2016\right)+x^3\left(x-2016\right)-x^2\left(x-2016\right)+x\left(x-2016\right)-\left(x-2016\right)+1\)

Thay x = 2016 vào ta được giá trị biểu thức trên = 1

Hok tốt!

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 ^,^