Question

simplificati 31 supra 3131,209 supra 209209 si ab supra abab, obtinand fractii ireductibile

Answer 1

   
[tex]\displaystyle \\ \frac{31}{3131} = \frac{31}{31 \times 101} = \boxed{\frac{1}{101} } \\ \\ \frac{209}{209209} = \frac{209}{209 \times 1001} = \boxed{\frac{1}{1001} } \\ \\ \frac{\overline{ab}}{\overline{abab}} = \frac{10a+b}{1000a+100b+10a+b} = \\\\=\frac{(10a+b)}{100(10a+b)+(10a+b)} = \\\\= \frac{(10a+b)}{(10a+b)(100+1)}= \frac{(10a+b)}{101(10a+b)}= \boxed{\frac{1}{101}}[/tex]