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Question-162027




Question Number 162027 by mathlove last updated on 25/Dec/21
Answered by MJS_new last updated on 25/Dec/21
x∈R∧y∈R∧n∈Z: y=((−x))^(1/(2n+1)) =−(x)^(1/(2n+1))   ⇒  ((−9^(−2) ))^(1/3) =−(9^(−2) )^(1/3) =−((1/9^2 ))^(1/3) =−(1/( ((81))^(1/3) ))=−(1/(3(3)^(1/3) ))
$${x}\in\mathbb{R}\wedge{y}\in\mathbb{R}\wedge{n}\in\mathbb{Z}:\:{y}=\sqrt[{\mathrm{2}{n}+\mathrm{1}}]{−{x}}=−\sqrt[{\mathrm{2}{n}+\mathrm{1}}]{{x}} \\ $$$$\Rightarrow \\ $$$$\sqrt[{\mathrm{3}}]{−\mathrm{9}^{−\mathrm{2}} }=−\sqrt[{\mathrm{3}}]{\mathrm{9}^{−\mathrm{2}} }=−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{81}}}=−\frac{\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}} \\ $$

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