Question

me ajudem por favor!​

Answer 1

Observe que:

$32 = 2\cdot2\cdot2\cdot2\cdot2= 2^5\\\\ 16 = 2\cdot2\cdot2\cdot2=2^4\\\\8 =2\cdot2\cdot2=2^3\\\\4 = 2\cdot2=2^2\\\\2=2\cdot1=2^1\\\\~\\ (a^b)^c = a^{b\cdot c} \\ \\~\\ ac+bc = c(a+b)$

Agora a questão:

$\dfrac{2^{98}+4^{50}-8^{34}}{2^{99}-32^{20}+2^{101}}\\\\~\\ \dfrac{2^{98}+(2^2)^{50}-(2^3)^{34}}{2^{99}-(2^5)^{20}+2^{101}}\\\\~\\ \dfrac{2^{98}+2^{100}-2^{102}}{2^{99}-2^{100}+2^{101}} \\\\~\\ \dfrac{2^{98}(1 +2^2 -2^4)}{2^{99}(1-2^1 +2^2)}\\\\~\\ \dfrac{(1 +4 -16)}{2^{99-98}(1-2 +4)} \\\\~\\ \dfrac{-11}{2\cdot3} \\\\\dfrac{-11}{6}\\\\ -1.8333333....$