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P(x) = x2016 - 2015x2015 - 2015x2014 - ... - 2015x2 - 2015x
<=> P(x) = x2016 - 2016x2015 + x2015 - 2016x2014 + x2014 - ... - 2016x2 + x2 - 2016x + x
<=> P(2016) = 20162016 - 2016.20162015 + 20162015 - 2016.20162014 + 20162014 -...- 2016.20162 + 20162 - 2016.2016 + 2016
<=> P(2016)=20162016 - 20162016 + 20162015 - 20162015 + 20162014 - ... - 20163 + 20162 - 20162 + 2016
<=> P(2016) = 2016
Vậy P(2016) = 2016
Ta có:
P(2016) = 20162016 - 2015 . 20162015 - 2015 . 20162014 -.....- 2015 . 20162 - 2015 . 2016 - 1
P(2016) = 20162016 - ( 2016 - 1 ) . 20162015 - ( 2016 -1 ) . 20162014 - ..... - ( 2016 - 1 ) . 20162 - ( 2016 - 1 ) . 2016 - 1
P(2016)= 20162016 - 20162016 + 20162015 - 20162015 + 20162014 - ..... - 20163 + 20162 - 20162 + 2016 - 1
P(2016) = 2016 - 1
P(2016) = 2015.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tim x thuoc Z de bieu thuc sau day dat gia tri lon nhat :
B= 2015x+1/2016x-2016 (x thuoc Z; x khac 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có x=2016 => x-1=2015
Thay vào ta được :
A=x^6 -(x-1)x^5 - (x-1)x^4 -(x-1)x^3 - (x-1)x^2 - (x-1)x -x
= x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x=0
Thay x=2016 vào biểu thức trên ta được:
\(A=x^6-\left(x-1\right).x^5-\left(x-1\right).x^4-\cdot\left(x-1\right).x^3-\left(x-1\right).x^2-\left(x-1\right).x-x\)
\(=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x\)
\(=0\)
Vậy x=2016 là nghiệm của đa thức .
![](https://rs.olm.vn/images/avt/0.png?1311)
1 : 2015x = - 1
2016 2015
<=> 1 . 1 = - 1
2016 2015x 2015
<=> 2016 . 2015x = - 2015
<=> x = - 2015 = - 1
2016. 2015 2016
vậy x = - 1/ 2016
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)