Решите , заранее 2^0*2^-1*16= х^3*х^4/х^-2= 27^2/3*32^2/5= - вопрос №202116342272332255772814 от Lexakek111 10.06.2021 02:26

Question

Математика: Решите , заранее 2^0*2^-1*16= х^3*х^4/х^-2= 27^2/3*32^2/5=

Lexakek111 · Answer

1) = 2^0*2^(-1) * 2^4 = 2^(0-1+4) = 2^3=8

2) = x^3 * x^4 * x^2 =x^(3+4+2) = x^9

3) = (3^3)^2 / (3 ^1 *( 2^5)^(2/5) ) =
= 3^(3*2) / ((3^1) * 2^(5 * 2/5))=
= 3^(6-1) / 2^2 = 3^5 / 2^2 =
=243 /4 = 60.75