Предмет: Алгебра, автор: sdghjgdd37

ПОМОГИТЕ ПОЖАЛУЙСТА ОЧЕНЬ СРОЧНО ПРОШУ ВАС УМОЛЯЮ ​

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Автор ответа: zinaidazina
0

(4^2)^0=1

4^2*4^0=16*1=16

4^{-1}*4^{-1}=\frac{1}{4}*\frac{1}{4}=\frac{1}{16}

4^{-1}=\frac{1}{4}

Ответ:  4^{-1}*4^{-1}< 4^{-1}< (4^2)^{0}<4^{2}*4^{0}

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