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Question Number 65077 by naka3546 last updated on 24/Jul/19

	Answered by MJS last updated on 25/Jul/19
	$(1) a+b+(c^2 −8c+14)(√(a+b−2))=1 (√(a+b−2))=t⇒t≥0 ⇒ a+b=t^2 +2 t^2 +(c^2 −8v+14)t+1=0 t=−((c^2 −8c+14)/2)±(((c−4)(√((c−2)(c−6))))/2) ∧t≥0 trying we find t<0∀c∈R\{4} c=4 ⇒ t=1 ⇒ b=3−a (2) −3a+15+(√(−a^2 +3a+4))=3 ⇒ a=4 ⇒ b=−1 so the only solution is a=4 b=−1 c=4$
	$(1)$ $a + b + (c^{2} - 8 c + 14) \sqrt{a + b - 2} = 1$ $\sqrt{a + b - 2} = t \Rightarrow t ⩾ 0$ $\Rightarrow a + b = t^{2} + 2$ $t^{2} + (c^{2} - 8 v + 14) t + 1 = 0$ $t = - \frac{c^{2} - 8 c + 14}{2} \pm \frac{(c - 4) \sqrt{(c - 2) (c - 6)}}{2} \land t ⩾ 0$ $trying we find t < 0 \forall c \in R ∖ {4}$ $c = 4 \Rightarrow t = 1 \Rightarrow b = 3 - a$ $(2)$ $- 3 a + 15 + \sqrt{- a^{2} + 3 a + 4} = 3$ $\Rightarrow a = 4 \Rightarrow b = - 1$ $so the only solution is$ $a = 4 b = - 1 c = 4$
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