Question Number 224043 by ghnaseri last updated on 15/Aug/25
$${f}\left({x}\right)=\left(\sqrt{{x}−\mathrm{5}}\right)^{\mathrm{0}} \\ $$$${Dom}_{{f}} =? \\ $$
Answered by Jyrgen last updated on 16/Aug/25
![0^0 is not defined z^0 =1∀z∈C\{0} z=re^(iθ) with r>0∧θ∈(−π; π] z^0 =r^0 e^(0iθ) =1×1=1 (√(x−5))∈C∀x∈R ⇒ ((√(x−5)))^0 =1∀x∈R\{5}](https://www.tinkutara.com/question/Q224044.png)
$$\mathrm{0}^{\mathrm{0}} \:{is}\:{not}\:{defined} \\ $$$${z}^{\mathrm{0}} =\mathrm{1}\forall{z}\in\mathbb{C}\backslash\left\{\mathrm{0}\right\} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:{with}\:{r}>\mathrm{0}\wedge\theta\in\left(−\pi;\:\pi\right] \\ $$$${z}^{\mathrm{0}} ={r}^{\mathrm{0}} \mathrm{e}^{\mathrm{0i}\theta} =\mathrm{1}×\mathrm{1}=\mathrm{1} \\ $$$$\sqrt{{x}−\mathrm{5}}\in\mathbb{C}\forall{x}\in\mathbb{R}\:\Rightarrow\:\left(\sqrt{{x}−\mathrm{5}}\right)^{\mathrm{0}} =\mathrm{1}\forall{x}\in\mathbb{R}\backslash\left\{\mathrm{5}\right\} \\ $$
Answered by saadww last updated on 16/Aug/25
$${A}=\sqrt{{Xa}+{Ya}} \\ $$