👤
Darabaoctavian
a fost răspuns

vreau si eu putin ajutor; a={[(2^5^2003^0+3^0^4^2003+7^1^5^2003):5^1^5^2003]:(2^2003:2^2000)}^2004 si b=[(2^2)^3-2003^0]:3^2^1^2004-(1^4^3^2003+4^1^2^2003). Calculati(a-b)^2003 Semnul ^=la putere

Răspuns :

KeyheysuntNicol
a={[(2^5+1+7):5]:2^3}^2004= {[(32+8):5]:8}^2004= [(40:5):8]^2004= (8:8)^2004= 1^2004= 1 a=1
b=(64-1):8-(1+4)= 63:9-5=7-5=2 b=2
(a-b)^2003 (1-2)^2003=-1^2003 = -1
cu plăcere !

Alte intrebari