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

\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-xz=b\\z^2-xy=c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\) \(\Rightarrow ax+by+cz=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)⋮\left(x+y+z\right)\)

Lời giải:

$f(x)=x^2+ax+b$

$f(f(x)+x)=[f(x)+x]^2+a[f(x)+x]+b$

$=f(x)^2+x^2+2xf(x)+af(x)+ax+b$

$=f(x)^2+2xf(x)+af(x)+f(x)$

$=f(x)[f(x)+2x+a+1]$

$=f(x)(x^2+ax+b+2x+a+1)$

$=f(x)[(x+1)^2+a(x+1)+b]=f(x)f(x+1)$

Thay $x=2019$ vô thì:

$f(f(2019)+2019)=f(2019).f(2020)$. Do đó tồn tại số $k=f(2019)+2019\in\mathbb{Z}$ thỏa mãn đkđb. 

Ta có đpcm.