\(B=2016^2+2017^2-2\\ B=2016^2-1+2017^2-1\\ B=\left(2016-1\right)\left(2016+1\right)+\left(2017-1\right)\left(2017+1\right)\\ B=2015.2017+2016.2018=A\)

Thanks bạn nhiều

đương nhiên là 2017 lớn hơn rồi

$\textbf{1.}$

Ta có $5^{2017}+5^{2015}=5^{2015}(5^2+1)$

$=5^{2015}\cdot26$

$=5^{2015}\cdot13\cdot2.$

Vì $13\mid13\cdot2$ nên $13\mid\left(5^{2017}+5^{2015}\right).$

Vậy $5^{2017}+5^{2015}$ chia hết cho $13.$

$\textbf{2.}$

Giả sử $a^{2014}+b^{2015}+c^{2016}\vdots6.$

Ta có $a^{2016}-a^{2014}=a^{2014}(a^2-1)$

$=a^{2014}(a-1)(a+1).$

Vì $a(a-1)(a+1)\vdots6$ nên $a^{2014}(a-1)(a+1)=a^{2013}\cdot a(a-1)(a+1)\vdots6.$

Suy ra $a^{2016}\equiv a^{2014}\pmod6.$

Tương tự, $b^{2017}-b^{2015}=b^{2015}(b^2-1)$

$=b^{2014}\cdot b(b-1)(b+1)\vdots6,$ nên $b^{2017}\equiv b^{2015}\pmod6.$

Lại có $c^{2018}-c^{2016}=c^{2016}(c^2-1)$

$=c^{2015}\cdot c(c-1)(c+1)\vdots6,$ nên $c^{2018}\equiv c^{2016}\pmod6.$

Cộng ba đồng dư trên, $a^{2016}+b^{2017}+c^{2018}\equiva^{2014}+b^{2015}+c^{2016}\equiv0\pmod6.$

Vậy $a^{2016}+b^{2017}+c^{2018}$ chia hết cho $6.$

\(10A=\dfrac{10^{2017}+1+9}{10^{2017}+1}=1+\dfrac{9}{10^{2017}+1}\)

\(10B=\dfrac{10^{2018}+10}{10^{2018}+1}=1+\dfrac{9}{10^{2018}+1}\)

Vì \(10^{2017}+1< 10^{2018}+1\)

nên A>B