Ta có:

$\left \lfloor \left ( \frac{1}{81}-\frac{3}{162} \right ).\frac{81}{17}+\frac{35}{34}\right \rfloor$

= $\left ( \frac{2}{162}-\frac{3}{162} \right ).\frac{81}{17}+ \frac{35}{34}$

=$\frac{-1}{162}.\frac{81}{17}+\frac{35}{34}$

=$\frac{-1}{34}+\frac{35}{34}=1$

$\left [ \left ( \frac{9}{51}+\frac{7}{102} \right ).\frac{102}{5}+2017 \right ]$

= $\left ( \frac{18}{102}-\frac{7}{102} \right ).\frac{102}{5}+ 2017$

=  $\frac{25}{102}.\frac{102}{5}+2017 = 2022$

A= $\frac{1}{2022}$