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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...
<3 bn nhìu nhìu
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
\(f\left(x\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-...-\left(x+1\right)x\)\(+x+2\)
\(=x^8-x^8-x^7+x^7+x^6-...+x^2-x^2-x+x+2\)
\(=2\)
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)
\(Q=1009\)
Tính giá trị của đa thức sau biết x=2018
N=x^6-2017x^5-2017x^4-2017x^3-2017x^2-2017x-2017
Help me :(((
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
\(f\left(x\right)=x^{99}-2017^{x^{98}}+2017^{x^{97}}-...+2017x-1\)
\(f\left(2016\right)=2016^{99}-2017.2016^{98}+2017.2016^{97}-...+2017.2016-1\)
\(f\left(2016\right)=2016^{99}-\left(2016+1\right).2016^{98}+\left(2016+1\right).2016^{97}-...+\left(2016+1\right).2016-1\)
\(f\left(2016\right)=2016^{99}-2016^{99}-2016^{98}+2016^{98}+2016^{97}-2016^{97}-2016^{96}+...+2016^2+2016-1\)
\(f\left(2016\right)=2016-1\)
\(f\left(2016\right)=2015\)