Question

помогите, буду благодарна)
(1,2,3)

Answer 1

$displaystylett1) frac{8^{11}-8^{10}-8^{9}}{4^{15}-8^{14}-8^{13}}= frac{8^{9}(8^2-8-1)}{4^{13}(4^2-4-1)}=frac{{(2^3)}^{9}(64-9)}{{(2^2)}^{13}(16-5)}=frac{2^{27}cdot55}{2^{26}cdot11}=\\{} =2cdot5=10 \\\ 2) frac{9^{23}-9^{22}-9^{21}}{27^{14}-27^{13}}=frac{9^{21}(9^{2}-9-1)}{27^{13}(27-1)}=frac{{(3^2)}^{21}(81-10)}{{(3^3)}^{13}cdot26}=frac{3^{42}cdot71}{3^{39}cdot26}=\\{} =frac{3^3cdot71}{26}=frac{27cdot71}{26}=frac{1917}{26}=73frac{19}{26}$

$displaystylett 3) frac{{(8^{n-2}+8^{n-3})}^2}{{(4^{n-1}-4^{n-2})}^3}=frac{{(8^{n-3}(8+1))}^2}{{(4^{n-2}(4-1))}^3}=frac{{({(2^3)}^{n-3}cdot9)}^2}{{({(2^2)}^{n-2}cdot3)}^3}=frac{{(2^{3n-9}cdot3^2)}^2}{{(2^{2n-4}cdot3)}^3}=\\{} =frac{{(2^{3n-9})}^2cdot{(3^2)}^2}{{(2^{2n-4})}^3cdot3^3}=frac{2^{6n-18}cdot3^4}{2^{6n-12}cdot3^3}=frac{3}{2^6}=frac{3}{64}$

Answer 2

спасибо большое))

Answer 3

пожалуйста)