Question Number 171043 by mathlove last updated on 06/Jun/22
$${A}\in{R} \\ $$$${A}=\frac{\sqrt{{x}−\mathrm{2}}+{x}+\mathrm{3}}{\:\sqrt{\mathrm{4}−\mathrm{2}{x}}+{x}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:{faind}\:{A}=? \\ $$
Answered by MJS_new last updated on 07/Jun/22
$$\sqrt{{x}−\mathrm{2}}\in\mathbb{R}\:\Rightarrow\:{x}\geqslant\mathrm{2} \\ $$$$\sqrt{\mathrm{4}−\mathrm{2}{x}}=\sqrt{\mathrm{2}}\sqrt{\mathrm{2}−{x}}\in\mathbb{R}\:\Rightarrow\:{x}\leqslant\mathrm{2} \\ $$$$\Rightarrow\:{x}=\mathrm{2} \\ $$$$\Rightarrow\:{A}=\mathrm{5} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Jun/22
$$\mathbb{G}\boldsymbol{\mathrm{reat}}\:\boldsymbol{\mathrm{sir}}! \\ $$
Commented by MJS_new last updated on 07/Jun/22
$$\mathrm{there}\:\mathrm{are}\:\mathrm{solutions}\:\mathrm{with}\:{x}\notin\mathbb{R}\wedge{A}\in\mathbb{R} \\ $$$$\mathrm{simply}\:\mathrm{set}\:{A}\in\mathbb{R}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$${A}=\mathrm{5}\:\Rightarrow\:{x}=\mathrm{2}\vee{x}=−\frac{\mathrm{17}}{\mathrm{16}}\pm\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{i} \\ $$
Commented by mathlove last updated on 08/Jun/22
$${thanks} \\ $$