Ta có:
\(A=\frac{2^{18}.18^7.3^3+3^{15}.2^{15}}{2^{10}.6^{15}+3^{14}.15.4^{13}}=\frac{2^{18}.\left(2.3^2\right)^7.3^3+3^{15}.2^{15}}{2^{10}.\left(2.3\right)^{15}+3^{14}.3.5.\left(2^2\right)^{13}}\)
\(=\frac{2^{18}.2^7.3^{14}.3^3+3^{15}.2^{15}}{2^{10}.2^{15}.3^{15}+3^{15}.5.2^{16}}=\frac{2^{25}.3^{17}+2^{15}.3^{15}}{2^{25}.3^{15}+3^{15}.2^{16}.5}=\frac{2^{15}.3^{15}.\left(3^2.2^{10}+1\right)}{2^{16}.3^{15}.\left(2^9+5\right)}\)
\(=\frac{3^2.2^{10}+1}{2^{10}+10}=\frac{9.1024+1}{1024+10}=\frac{9217}{1025}\)