Давай упростим выражение, а потом подставим значение k.
\(\frac{5(2k)^4}\){k^{17}k^5} = \(\frac{5 \times 2^4 \times k^4}\){k^{17+5}} = \(\frac{5 \times 16 \times k^4}\){k^{22}} = \(\frac{80}\){k^{22-4}} = \(\frac{80}\){k^{18}}k = 2√5. Теперь возведем это в 18-ю степень:k^{18} = (2√5)^{18} = 2^{18} \(\times\) (√5)^{18} = 2^{18} \(\times\) 5^9\(\frac{80}\){k^{18}} = \(\frac{80}\){2^{18} \(\times\) 5^9}80 = 16 \(\times\) 5 = 2^4 \(\times\) 5.\(\frac{2^4 \times 5}\){2^{18} \(\times\) 5^9} = \(\frac{1}\){2^{18-4} \(\times\) 5^{9-1}} = \(\frac{1}\){2^{14} \(\times\) 5^8}Ответ: \(\frac{1}\){2^{14} \(\times\) 5^8}