Question

am 2 probleme:
1. stabileste daca 1 supra 3 + 1 supra 6 +1 supra 12 mai mic decat 1
2.afla a din egalitatea( 1 supra 3 + 1 supra 6+1 supra 12) + a=1 va rog e urgent

Answer 1

$\frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4+2+1}{12} => \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{7}{12}$
$1 < 7 \ | : 12 => \frac{1}{12} < \frac{7}{12} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12}$
In consecinta: $\frac{1}{3} + \frac{1}{6} + \frac{1}{12} >1$
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2)   $\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + a =1 <=> \frac{7}{12} + a =1 =>$
$a = 1- \frac{7}{12} <=> a = \frac{12-7}{12} => a = \frac{5}{12}$
Asadar:  $a = \frac{5}{12}$