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Question Number 170729 by solomonwells last updated on 30/May/22

Commented by phamkhanhhuong last updated on 30/May/22

$\Leftrightarrow \sqrt{x - 1 + 2 \sqrt{x - 1} + 1} + \sqrt{x - 1 - 2 \sqrt{x - 1} + 1} = x - 1$ $\Leftrightarrow \sqrt{{(\sqrt{x - 1} + 1)}^{2}} + \sqrt{{(\sqrt{x - 1} - 1)}^{2}} = x - 1$ $\Leftrightarrow∣ \sqrt{x - 1} + 1 ∣ + ∣ \sqrt{x - 1} - 1 ∣= x - 1 ()$ $\sqrt{x - 1} + 1 > 0 \forall x \in [1; + \infty) .$ $∙ \sqrt{x - 1} ⩾ 1 \Leftrightarrow x ⩾ 2$ $() \Leftrightarrow 2 \sqrt{x - 1} = x - 1 \Leftrightarrow \sqrt{x - 1} = 0 o r 2 = \sqrt{x - 1}$ $\Leftrightarrow x = 5 .$ $∙ \sqrt{x - 1} < 1 \Leftrightarrow x < 2$ $(*) \Leftrightarrow 2 = x - 1 \Leftrightarrow x = 3 (. . .)$ $K L : x = 5$
Commented by Rasheed.Sindhi last updated on 30/May/22

$⋒ \overset{∙}{∥} ⋐ \overset{}{\in}!$
	Answered by MJS_new last updated on 30/May/22

	$\sqrt{(x - 1) + 1 + 2 \sqrt{x - 1}} + \sqrt{(x - 1) + 1 - 2 \sqrt{x - 1}} = x - 1$ $\sqrt{x - 1} = t \land t ⩾ 0$ $\sqrt{t^{2} + 2 t + 1} + \sqrt{t^{2} - 2 t + 1} = t^{2}$ $∣ t + 1 ∣ + ∣ t - 1 ∣= t^{2}$ $(1) 0 ⩽ t ⩽ 1 \Rightarrow ∣ t - 1 ∣= 1 - t$ $2 = t^{2} \Rightarrow t = \sqrt{2} > 1 no solution$ $(2) t > 1 \Rightarrow ∣ t - 1 ∣= t - 1$ $2 t = t^{2} \Rightarrow (t = 0 < 1 no solution) \lor t = 2$ $\Rightarrow$ $t = \sqrt{2} \Rightarrow x - 1 = 4 \Rightarrow x = 5$
	Answered by som(math1967) last updated on 30/May/22
	${(√(x+2(√(x−1))))+(√(x−2(√(x−1))))}^2 =(x−1)^2 ⇒x+2(√(x−1))+x−2(√(x−1))+2(√(x^2 −4x+4))=(x−1)^2 ⇒2x+2(x−2)=x^2 −2x+1 ⇒x^2 −6x+5=0 ⇒(x−5)(x−1)=0 x=5$
	${\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}}}^{2} = {(x - 1)}^{2}$ $\Rightarrow x + 2 \sqrt{x - 1} + x - 2 \sqrt{x - 1} + 2 \sqrt{x^{2} - 4 x + 4} = {(x - 1)}^{2}$ $\Rightarrow 2 x + 2 (x - 2) = x^{2} - 2 x + 1$ $\Rightarrow x^{2} - 6 x + 5 = 0$ $\Rightarrow (x - 5) (x - 1) = 0$ $x = 5$
	Commented by Tawa11 last updated on 30/May/22

	$Great sir$
	Answered by Rasheed.Sindhi last updated on 30/May/22

	$\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} = x - 1$ $M u l t i p l y i n g b y \sqrt{x - 2 \sqrt{x - 1}} :$ $\sqrt{x^{2} - 4 (x - 1)} + x - 2 \sqrt{x - 1} = (x - 1) \sqrt{x - 2 \sqrt{x - 1}}$ $2 x - 2 - 2 \sqrt{x - 1} = (x - 1) \sqrt{x - 2 \sqrt{x - 1}}$ $2 (x - 1 - \sqrt{x - 1}) = (x - 1) \sqrt{x - 2 \sqrt{x - 1}}$ $\sqrt{x - 1} = y \Rightarrow x - 1 = y^{2} \Rightarrow x = y^{2} + 1$ $2 (y^{2} - y) = y^{2} \sqrt{y^{2} + 1 - 2 y}$ $2 (y^{2} - y) = y^{2} (y - 1)$ $2 y (y - 1) - y^{2} (y - 1) = 0$ $y (y - 1) (2 - y) = 0$ $y = 0 ∣ y - 1 = 0 ∣ y - 2 = 0$ $∙ \sqrt{x - 1} = 0 \Rightarrow x = 0 (N o t v a l i d)$ $∙ \sqrt{x - 1} - 1 = 0 \Rightarrow x - 1 = 1 \Rightarrow x = 2 (N o t v a l i d)$ $∙ \sqrt{x - 1} - 2 = 0 \Rightarrow x - 1 = 4 \Rightarrow x = 5 ✓$
	Commented by Tawa11 last updated on 30/May/22

	$Great sir$
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