очень прошу о помощи. Created by meow75. algebra-ru

очень прошу о помощи...

July 2022 1 2 Report

очень прошу о помощи

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ytekjdbvsq (2^(1/3) · 2^(1/4) : 2^(1/12) )² = 2^(2/3) · 2^(2/4) : 2^(2/12) = 2^(2/3) · 2^(1/2) : 2^(1/6) = 2^(2/3 + 1/2 – 1/6) = 2^((4+3–1)/6) = 2^(6/6) = 2^1 = 2

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