Question

calculati:
[1,3la puterea 4 :1,3²:1,3+1,3³la puterea0*(3,1²-3²)*10²]:3,1=

Answer 1

[1,3^4:1,3^2:1,3+(1,3^3)^0×(3,1^2-3^2)×10^2]:3,1=
[1,3^2:1,3+1×(9,61-9)×100]:3,1=
(1,3+0,61×100):3,1=
(1,3+61):3,1=
62,3:3,1=20,0967742

Answer 2

[1,3⁴:1,3²:1,3+1,3³la puterea 9×(3,1²-0²)×10²]:3,1=
1,3⁴-(scazi doar la putere)1,3²:1,3+1,3la puterea0×(9,61-9)×10²]:3,1=
(1,3²:1,3+1×0,61×100):3,1=
(1,3+1,61×100):3,1=
(0,31×100):3,1=
31:3,1=
31(×10):3,1(×10)=
310:31=
10

Sper ca te-am ajutat!