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Question Number 109963 by bobhans last updated on 26/Aug/20

	Answered by bemath last updated on 26/Aug/20
	$(1)(√(x^2 −2x+2)) = (√((x−1)^2 +1)) ≥ 1 (2)(√(x^2 −2x+10)) =(√((x−1)^2 +9)) ≥ 3 ⇔ log _3 (√((x−1)^2 +9)) ≥ 1 therefore (√(x^2 −2x+2)) + log _3 (√(x^2 −2x+10)) =2 where { (((√(x^2 −2x+2)) =1 ∧)),((log _3 (√(x^2 −2x+10)) = 1)) :} we get x = 1$
	$(1) \sqrt{x^{2} - 2 x + 2} = \sqrt{{(x - 1)}^{2} + 1} ⩾ 1$ $(2) \sqrt{x^{2} - 2 x + 10} = \sqrt{{(x - 1)}^{2} + 9} ⩾ 3$ $\Leftrightarrow \log_{3} \sqrt{{(x - 1)}^{2} + 9} ⩾ 1$ $t h e r e f o r e$ $\sqrt{x^{2} - 2 x + 2} + \log_{3} \sqrt{x^{2} - 2 x + 10} = 2$ $w h e r e {\begin{cases} \sqrt{x^{2} - 2 x + 2} = 1 \land \\ \log_{3} \sqrt{x^{2} - 2 x + 10} = 1 \end{cases}$ $w e g e t x = 1$
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