Question

поиогите
огэ номер 21

Answer 1

=27y^3 * y^-10/y^-12 * 2y^5 = 27y^-27/2y^_7=13,5

Answer 2

$\frac{(3y)^3*y^{-10}}{y^{-12}*2y^5}=\frac{27y^3*y^{12}}{y^{10}*2y^5}=\frac{27y^{15}}{2y^{15}}=\frac{27}{2}=13,5$