Question

3^22-3×(3^2+3^3)+6^0​

Answer 1

Răspuns:

$\bf 3^{22}-3\cdot (3^2+3^3)+6^0=$

$\bf 3^{22}-3\cdot (9+27)+1=$

$\bf 3^{22}-3\cdot 36+1= 3^{22}-3\cdot 36+1=$

$\bf 3^{22}-108+1=\red{\underline{~3^{22}-107~}}$