Question Number 144860 by mathdanisur last updated on 29/Jun/21
$$\mid\frac{{x}}{{x}-\mathrm{1}}\mid\:+\:\mid{x}\mid\:=\:\frac{{x}^{\mathrm{2}} }{\mid{x}-\mathrm{1}\mid}\:\:\:{find}\:\:{x}=? \\ $$
Commented by hknkrc46 last updated on 29/Jun/21
$$\bigstar\:\mid\boldsymbol{{x}}\mid\:=\:\frac{\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mid\boldsymbol{{x}}\mid}{\mid\boldsymbol{{x}}\:−\:\mathrm{1}\mid} \\ $$$$\bigstar\:\mid\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}}\mid\:=\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mid\boldsymbol{{x}}\mid \\ $$$$\:\:\:\left(\boldsymbol{{a}}\right)\:\boldsymbol{{x}}\:\in\:\left[\mathrm{0},\mathrm{1}\right)\:\Rightarrow\:\boldsymbol{{x}}\:−\:\boldsymbol{{x}}^{\mathrm{2}} \:=\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\boldsymbol{{x}}\:=\:\mathrm{0} \\ $$$$\:\:\:\left(\boldsymbol{{b}}\right)\:\boldsymbol{{x}}\:\in\:\mathbb{R}^{−} \:\Rightarrow\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}}\:=\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\boldsymbol{{x}}\:=\:\mathrm{0}\:\notin\:\mathbb{R}^{−} \\ $$$$\:\:\:\left(\boldsymbol{{c}}\right)\:\boldsymbol{{x}}\:\in\:\left(\mathrm{1}\:,\:\infty\right)\:\Rightarrow\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}}\:=\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{0}\:=\:\mathrm{0}\: \\ $$$$\clubsuit\:\:\boldsymbol{{x}}\:\in\:\left(\mathrm{1}\:,\:\infty\right)\:\cup\:\left\{\mathrm{0}\right\} \\ $$
Commented by mathdanisur last updated on 30/Jun/21
$${alot}\:{cool}\:{Sir}\:{thanks} \\ $$