Question

10[(3-6)+(4/2)/ [6-8(3-6)]+2(6)+1
ayuden. ​

Answer 1

Primero se resuelve dentro de paréntesis, divisiones y multiplicaciones

$10[(3-6)+(4/2)/[6-8(3-6)]+2(6)+1]\\10[(-3)+(2)/[6-8(-3)]+2(6)+1]\\10[-3+2/(6+24)+12+1]\\10[-3+2/30+12+1]\\10(-3+1/15+12+1)\\10(-3+1/15+13)\\10(1/15+10)\\10(\frac{151}{15} )\\\\\frac{1510}{15}=\frac{302}{15}$