Răspuns :
3x[3+3x(3^3-3^5:3^3)]+2x3^2
=
3x[3+3x(3^3-3^2)]+2x3^2 =
3x[3+3x3^2x(3-1)]+2x3^2 =
3x[3+3^3x2]+2x3^2 =
3^2+3^4x2+2x3^2 =
3x3^2+2x3^4=
3^3+2x3^4=
3^3x(1+2x3)=
3^3x7=
27x7=
189
3x[3+3x(3^3-3^2)]+2x3^2 =
3x[3+3x3^2x(3-1)]+2x3^2 =
3x[3+3^3x2]+2x3^2 =
3^2+3^4x2+2x3^2 =
3x3^2+2x3^4=
3^3+2x3^4=
3^3x(1+2x3)=
3^3x7=
27x7=
189