∣(x/(x-1))∣ + ∣x∣ = (x^2 /(∣x-1∣)) find x=?


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Question Number 144860 by mathdanisur last updated on 29/Jun/21

$∣ \frac{x}{x - 1} ∣ + ∣ x ∣ = \frac{x^{2}}{∣ x - 1 ∣} f i n d x = ?$
Commented by hknkrc46 last updated on 29/Jun/21
$★ ∣x∣ = ((x^2 − ∣x∣)/(∣x − 1∣)) ★ ∣x^2 − x∣ = x^2 − ∣x∣ (a) x ∈ [0,1) ⇒ x − x^2 = x^2 − x ⇒ x = 0 (b) x ∈ R^− ⇒ x^2 − x = x^2 + x ⇒ x = 0 ∉ R^− (c) x ∈ (1 , ∞) ⇒ x^2 − x = x^2 − x ⇒ 0 = 0 ♣ x ∈ (1 , ∞) ∪ {0}$
$★ ∣ x ∣ = \frac{x^{2} - ∣ x ∣}{∣ x - 1 ∣}$ $★ ∣ x^{2} - x ∣ = x^{2} - ∣ x ∣$ $(a) x \in [0, 1) \Rightarrow x - x^{2} = x^{2} - x$ $\Rightarrow x = 0$ $(b) x \in R^{-} \Rightarrow x^{2} - x = x^{2} + x$ $\Rightarrow x = 0 \notin R^{-}$ $(c) x \in (1, \infty) \Rightarrow x^{2} - x = x^{2} - x$ $\Rightarrow 0 = 0$ $♣ x \in (1, \infty) \cup {0}$
Commented by mathdanisur last updated on 30/Jun/21

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