Tích

A

- Hướng dẫn giải

Đáp án A.

$\left(2017!\right){\left(1+\frac{1}{1}\right)}^1{\left(1+\frac{1}{2}\right)}^2\mathrm{...}{\left(1+\frac{1}{2017}\right)}^{2017}$$=\left(2017!\right){\left(\frac{2}{1}\right)}^1{\left(\frac{3}{2}\right)}^2{\left(\frac{4}{3}\right)}^3\mathrm{...}{\left(\frac{2018}{2017}\right)}^{2017}$

$=\left(2017!\right).\frac{2^1{.3}^2{.4}^3{\mathrm{....2018}}^{2017}}{1^1{.2}^2{.3}^3{\mathrm{...2017}}^{2017}}$$=\left(2017!\right).\frac{{2018}^{2017}}{\mathrm{1.2.3...2017}}$$=2017!.\frac{{2018}^{2017}}{2017!}={2018}^{2017}$

Suy ra $a=2018;b=2017$