If { ((a+(1/b)=(7/3))),((b+(1/c)=4)),((c+(1/a)=1)) :} find 2022∙abc


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Question Number 184431 by SulaymonNorboyev last updated on 06/Jan/23

$I f {\begin{cases} a + \frac{1}{b} = \frac{7}{3} \\ b + \frac{1}{c} = 4 \\ c + \frac{1}{a} = 1 \end{cases} find 2022 \cdot abc$
	Answered by mr W last updated on 06/Jan/23

	$c = 1 - \frac{1}{a} = \frac{a - 1}{a}$ $b = 4 - \frac{a}{a - 1} = \frac{3 a - 4}{a - 1}$ $\Rightarrow a b c = a \times \frac{3 a - 4}{a - 1} \times \frac{a - 1}{a} = 3 a - 4$ $a + \frac{a - 1}{3 a - 4} = \frac{7}{3}$ $9 a^{2} - 30 a + 25 = 0$ ${(3 a - 5)}^{2} = 0$ $\Rightarrow a = \frac{5}{3}$ $a b c = 3 a - 4 = 5 - 4 = 1$ $2023 a b c = 2023 ✓$
	Answered by ajfour last updated on 06/Jan/23
	$(a+(1/b))(b+(1/c))(c+(1/a))=pqr abc+b+a+(1/c)+c+(1/a)+(1/b)+(1/(abc)) =pqr abc+(1/(abc))+(p+q+r)=pqr (abc)((1/(abc)))=1 ⇒ abc, (1/(abc))=((pqr−(p+q+r))/2) ±(√({((pqr−(p+q+r))/2)}^2 −1)) =((14)/3)−((11)/3)±(√(1−1)) =1$
	$(a + \frac{1}{b}) (b + \frac{1}{c}) (c + \frac{1}{a}) = p q r$ $a b c + b + a + \frac{1}{c} + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{a b c}$ $= p q r$ $a b c + \frac{1}{a b c} + (p + q + r) = p q r$ $(a b c) (\frac{1}{a b c}) = 1$ $\Rightarrow a b c, \frac{1}{a b c} = \frac{p q r - (p + q + r)}{2}$ $\pm \sqrt{{\frac{p q r - (p + q + r)}{2}}^{2} - 1}$ $= \frac{14}{3} - \frac{11}{3} \pm \sqrt{1 - 1} = 1$
	Commented by mr W last updated on 06/Jan/23

	$v e r y s m a r t a p p r o a c h!$
	Commented by manxsol last updated on 07/Jan/23

	$(a b c) (\frac{1}{a b c}) = 1 t o t h e t o o l b o x$
	Commented by manxsol last updated on 07/Jan/23

	$a s u r r e a l a n s w e r$
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