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19to the power of 98 +(342)(19 to the power of 97) / 19 to the power of 99=?

$\frac{(19)^{98} + 342 (19)^{97}}{(19)^{99}} = \frac{19^{98} + 18 \times 19 (19)^{97}}{19^{99}} = \frac{19^{98} + 18 \times 19^{97 + 1}}{19^{99}} {∵ a^{p} . a^{q} = a^{p + q}} = \frac{19^{98} (1 + 18)}{19^{99}} = \frac{19^{98} (19)}{19^{99}} = \frac{19^{99}}{19^{99}} = 19^{99 - 99} = 19^{0} = 1$

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