Question

[Nguyễn Nhật Minh]

Cho đa thức P(x) = ax + b với a,b thuộc Z; a khác 0.

Chứng minh rằng |P(2018) - P(1)| \(\ge\)2017

Answer 1

[Arima Kousei]

Ta có : 

\(P\left(x\right)=ax+b\)

\(\Rightarrow\hept{\begin{cases}P\left(2018\right)=a.2018+b\\P\left(1\right)=a.1+b\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}P\left(2018\right)=2018a+b\\P\left(1\right)=a+b\end{cases}}\)

\(\Rightarrow P\left(2018\right)-P\left(1\right)=2018a+b-\left(a+b\right)\)

\(\Rightarrow P\left(2018\right)-P\left(1\right)=2017a\)

\(\Rightarrow\left|P\left(2018\right)-P\left(1\right)\right|=\left|2017a\right|\)

Do a khác 0 

\(\Rightarrow\left|2017a\right|\ge2017\)

\(\Rightarrow\left|P\left(2018\right)-P\left(1\right)\right|\ge2017\)

Vậy \(\left|P\left(2018\right)-P\left(1\right)\right|\ge2017\left(đpcm\right)\)