Question

va rog mult h si i dau coroana​

Answer 1

Răspuns:

h) [(9+8): 17]^2017- 0

(17:17)^2017

1^2017

=1

i)

{1+[2+(3+1024): 79]:5}^5 :512 -2

[1+(2 + 1027:79):5]^5 :512-2

[1+(2+13):5]^5:512-2

(1+15:5)^5:512-2

(1+3)^5:512-2

1024:512-2

2-2

=0

Answer 2

Răspuns:

h) [(3^2+2^3):17]^2017-0^2017=

[(9+8):17]^2017-0^2017=

(17:17)^2017-0^2017=

1^2017-0^2017=

1-0=1

i) {1+[2+(3+4^5):79]:5}^5:512-2=

{1+[2+(3+1024):79]:5}^5:512-2=

4^5=4•4•4•4•4=16•16•4=256•4=1024

[1+(2+1027:79):5]^5:512-2=

[1+(2+13):5]^5:512-2=

(1+15:5)^5:512-2

(1+3)^5:512-2=

4^5:512-2=

1024:512-2=

2-2=0

^=la puterea

sper că te-am ajutat!

coroniță?!