Question

daca nr. naturale prime a,b,c,a<b<c,sunt astfel incat 1/a+1/b+1/c=41/42, atunci sumaa+b+ceste egala cu:​

Answer 1

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{41}{42}$

Considerăm:

a=2 și b=3

$\frac{1}{2} + \frac{1}{3} + \frac{1}{c} = \frac{41}{42}$

$\frac{21}{42} + \frac{14}{42} + \frac{ 1}{c} = \frac{41}{42}$

$41 - 21 - 14 = 6 = > \frac{1}{c} = \frac{6}{42}$

$c = \frac{42}{6} = 7$

Răspuns: a=2 , b=3, c=7