Đáp án:

$\begin{array}{l}
B1)\\
A = \dfrac{{2016.2017 - 1}}{{2016.2017}}\\
 = \dfrac{{2016.2017}}{{2016.2017}} - \dfrac{1}{{2016.2017}}\\
 = 1 - \dfrac{1}{{2016.2017}}\\
B = \dfrac{{2017.2018 - 1}}{{2017.2018}}\\
 = 1 - \dfrac{1}{{2017.2018}}\\
Do:2016.2017 < 2017.2018\\
 \Rightarrow \dfrac{1}{{2016.2017}} > \dfrac{1}{{2017.2018}}\\
 \Rightarrow  - \dfrac{1}{{2016.2017}} <  - \dfrac{1}{{2017.2018}}\\
 \Rightarrow 1 - \dfrac{1}{{2016.2017}} < 1 - \dfrac{1}{{2017.2018}}\\
 \Rightarrow A < B\\
B2)A = \dfrac{{2016.2017 + 1}}{{2016.2017}}\\
 = \dfrac{{2016.2017}}{{2016.2017}} + \dfrac{1}{{2016.2017}}\\
 = 1 + \dfrac{1}{{2016.2017}}\\
B = \dfrac{{2017.2018 + 1}}{{2017.2018}}\\
 = 1 + \dfrac{1}{{2017.2018}}\\
Do:2016.2017 < 2017.2018\\
 \Rightarrow \dfrac{1}{{2016.2017}} > \dfrac{1}{{2017.2018}}\\
 \Rightarrow 1 + \dfrac{1}{{2016.2017}} > 1 + \dfrac{1}{{2017.2018}}\\
 \Rightarrow A > B\\
B3)\\
A = \dfrac{{2016}}{{2017}} + \dfrac{{2017}}{{2018}}\\
B = \dfrac{{2016 + 2017}}{{2017 + 2018}}\\
 = \dfrac{{2016}}{{2017 + 2018}} + \dfrac{{2017}}{{2017 + 2018}}\\
Do:\dfrac{{2016}}{{2017}} > \dfrac{{2016}}{{2017 + 2018}}\\
\dfrac{{2017}}{{2018}} > \dfrac{{2017}}{{2017 + 2018}}\\
 \Rightarrow \dfrac{{2016}}{{2017}} + \dfrac{{2017}}{{2018}} > \dfrac{{2016}}{{2017 + 2018}} + \dfrac{{2017}}{{2017 + 2018}}\\
 \Rightarrow A > B\\
B4)\\
A = \dfrac{{{{17}^{18}} + 1}}{{{{17}^{19}} + 1}} < 1\\
 \Rightarrow \dfrac{{{{17}^{18}} + 1}}{{{{17}^{19}} + 1}} < \dfrac{{{{17}^{18}} + 1 + 16}}{{{{17}^{19}} + 1 + 16}}\\
 \Rightarrow A < \dfrac{{{{17}^{18}} + 17}}{{{{17}^{19}} + 17}}\\
 \Rightarrow A < \dfrac{{17\left( {{{17}^{17}} + 1} \right)}}{{17\left( {{{17}^{18}} + 1} \right)}}\\
 \Rightarrow A < \dfrac{{{{17}^{17}} + 1}}{{{{17}^{18}} + 1}}\\
 \Rightarrow A < B
\end{array}$