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

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:

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

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

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

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

\(x^6-2017x^5+2017x^4-2017x^3+2017x^2-2017x+2017\) (1)

Thay 2017 = x+1 vào  (1) ,có :

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

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

= 1

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)

\(Q=1009\)

Dễ thầy 2017=2016+1=x+1

Thay vào ta có:

\(x^{10}-2017x^9+2017x^8-.....+2017x^2-2017x+2017\)

\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-....+\left(x+1\right)x^2-\left(x+1\right)x+2017\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-....+x^3+x^2-x^2-x+2017=-x+2017=-2016+2017=1\)

Vậy..........

thanks bn!!

456545756858768978087

\(A=x^3+2x^2+3x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ =1999.\left(1999+1\right)=1999.2000\\ =3998000\)

\(B=x^4-2017x^3+2017x^2-2017x+2018\\ =x^4-2016x^3-x^3+2016x^2+x^2-2016x-x+2016+2\\ =x^3\left(x-2016\right)-x^3\left(x-2016\right)+x\left(x-2016\right)-\left(x-2016\right)+2\\ =\left(x-2016\right)\left(x^3+x-1\right)+2=0+2=0\)

Bạn xem lại đề câu a nhé , theo mk thì phải là 2 thì tính ms nhanh đc, 3 thì cũng giải đc nhưng ko hợp lí lắm

\(56:54+23.22-12017\)

\(=56.\dfrac{1}{54}+506-12017\)

\(=\dfrac{28}{27}-11511\)

\(=-11509,96296\)