Question Number 150435 by mathdanisur last updated on 12/Aug/21
$$\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mid\mathrm{x}\:-\:\mathrm{3}\mid^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:-\:\mathrm{8x}\:+\:\mathrm{15}}{\boldsymbol{\mathrm{x}}\:-\:\mathrm{2}}} \:=\:\mathrm{1} \\ $$
Answered by amin96 last updated on 12/Aug/21
$${x}\neq\mathrm{2}\:\:\:\Rightarrow\:\:{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}=\mathrm{0}\:\:\left({x}−\mathrm{5}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\begin{cases}{{x}_{\mathrm{1}} =\mathrm{3}\:\:\:\:\Rightarrow\:\:\mathrm{0}^{\mathrm{0}} =?}\\{{x}_{\mathrm{2}} =\mathrm{5}\:\:\checkmark\:\:\:\mid{x}−\mathrm{3}\mid=\mathrm{1}\:\:\:\Rightarrow\:\:\begin{cases}{{x}_{\mathrm{3}} =\mathrm{4}}\\{{x}_{\mathrm{4}} =\mathrm{2}\:\:\mathrm{x}\neq\mathrm{2}}\end{cases}}\end{cases}\:\:\:\: \\ $$$$\left(\mathrm{4};\:\:\mathrm{5}\right) \\ $$
Commented by mathdanisur last updated on 12/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$