Câu hỏi của Nguyễn Hương

Question

Cho P(x) = $x^{10}-2019x^9+2019x^8-2018x^7+...+2019$

Tính P(2018)

Nguyễn Khang · Accepted Answer

Ta có 2019=2018+1=x+1

Thay 2019=x+1 vào đa thức P(x) ta có :

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

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

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

$P\left(x\right)=x+1=2018+1=2019$Câu trả lời: Ta có 2019=2018+1=x+1

Thay 2019=x+1 vào đa thức P(x) ta có :

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

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

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

$P\left(x\right)=x+1=2018+1=2019$

Đặng Thu Hiền · Answer

Theo đề bài ta có 2019=2018+1=x+1

Thay 2019=x+1 vào đa thức P(x) ta có :

P(x)=x10−(x+1)x9+(x+1)x8−.......+(x+1)P(x)=x10−(x+1)x9+(x+1)x8−.......+(x+1)

P(x)=x10−x10−x9+x9+x8−.......+x+1P(x)=x10−x10−x9+x9+x8−.......+x+1

P(x)=(x10−x10)−(x9−x9)+(x8−x8)−....+x+1P(x)=(x10−x10)−(x9−x9)+(x8−x8)−....+x+1

P(x)=x+1=2018+1=2019Câu trả lời: Theo đề bài ta có 2019=2018+1=x+1

Thay 2019=x+1 vào đa thức P(x) ta có :

P(x)=x10−(x+1)x9+(x+1)x8−.......+(x+1)P(x)=x10−(x+1)x9+(x+1)x8−.......+(x+1)

P(x)=x10−x10−x9+x9+x8−.......+x+1P(x)=x10−x10−x9+x9+x8−.......+x+1

P(x)=(x10−x10)−(x9−x9)+(x8−x8)−....+x+1P(x)=(x10−x10)−(x9−x9)+(x8−x8)−....+x+1

P(x)=x+1=2018+1=2019

Chương III : Thống kê