Question

35 баллов,помогите!​

Answer 1

Ответ:

$6)\ \ \dfrac{3^7\cdot 27}{(3^4)^3}=\dfrac{3^7\cdot 3^3}{3^{12}}=\dfrac{3^{10}}{3^{12}}=\dfrac{1}{3^2}=\dfrac{1}{9}\\\\\\\dfrac{(7^3)^3\cdot 7^2}{(7^5)^2}=\dfrac{7^9\cdot 7^2}{7^{10}}=\dfrac{7^{11}}{7^{10}}=7^1=7\\\\\\7)\ \ y=-1\dfrac{2}{3}\ \ ,\ \ -2,5y^0=-2,5\cdot 1=-2,5\\\\\\a=\dfrac{2}{3}\ ,\ \ c=-\dfrac{1}{3}\ \ ,\ \ \ 27a^0c^3=27\cdot 1\cdot \Big(-\dfrac{1}{3}\Big)^3=27\cdot \Big(-\dfrac{1}{27}\Big)=-1$