\(\frac{5\cdot\left(3\cdot7^{15}-19\cdot7^{14}\right)}{7^{16}+3\cdot7^{14}}=\frac{5\cdot\left(3\cdot7-19\right)\cdot7^{14}}{\left(7^2+3\right)\cdot7^{14}}=\frac{5\cdot2\cdot7^{14}}{52\cdot7^{14}}=\frac{5\cdot2}{52}=\frac{5\cdot2}{26\cdot2}=\frac{5}{26}\)