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Question Number 165969 by leicianocosta last updated on 10/Feb/22

Commented by leicianocosta last updated on 10/Feb/22

Commented by hknkrc46 last updated on 10/Feb/22

$$▸{(-3)}^0=1\because \mathrm{\forall }\mathit{x}\in {\mathbb{R}}^{\ast }\Rightarrow {\mathit{x}}^0=1$$$$({\mathbb{R}}^{\ast }=\mathbb{R}-\left\{0\right\})$$$$▸-{(-3)}^3=-(-3^3)=-(-27)=27$$$$\because 2\nmid \mathit{a}\wedge \mathit{b}\in {\mathbb{R}}^{+}\Rightarrow {(-\mathit{b})}^{\mathit{a}}=-{\mathit{b}}^{\mathit{a}}$$$$▸{(-2)}^{-1}=\frac{1}{{(-2)}^1}=\frac{1}{-2^1}=-\frac{1}{2}$$$$▸-\left(5^{-2}\right)=-\frac{1}{5^2}=-\frac{1}{25}$$$$★\mathit{A}=\frac{1\cdot 27\cdot 5}{(-\frac{1}{2})(-\frac{1}{25})}=6750$$

Commented by alephzero last updated on 11/Feb/22

$$A=\frac{{(-3)}^0(-{(-3)}^3)5}{{(-2)}^{-1}(-\left(5^{-2}\right))}=$$$$=\frac{5\left(27\right)}{\frac{1}{2\left(25\right)}}=5\cdot 27\cdot 2\cdot 25=270\cdot 25$$$$=6750$$

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