Question

Calculati,respectand ordinea efectuarii operatiilor:

a) $1.2^{2} + 0.4^{2} / 0.2^{3}$
b) 0.2/(6.3-3.51/2.7)
c) 9.1*10-8.9*(10-0.5)
d) (7.8-2.3)*(7.8+2.3)

Answer 1

a)1,2=12/10=6/5; 0,4=4/10=2/5; 0,2=2/10=1/5
  (6/5)*(6/5)+[(2/5)*(2/5)]/[(1/5)*(1/5)*(1/5)]=
  36/25+(4/25)(/1/125)=  //se va transforma in inmultire: (4*125)/25
  36/25+20= //amplificam cu 25
  (36+500)/25=536/25  
 
b)(2/10)/[63/10-(351/100)/(27/10)]=
   (2/10)/(63/10-351/270)= //simplificam (351/270) cu 9
   (2/10)/(63/10-39/30)= //ampificam (63/10) cu 3
   (2/10)/[(189-39)/30]=
   (2/10)/(150/30)=
   (2/10)/5=
   (1/5)/5=1/25

c)91-89- 89* (-1/2)= //  -0,5=-1/2
   2+89/2= // amplificam cu 2
   (4+89)/2=93/2
 
d) Folosim formula (a+b)(a-b)=a*a-b*b
   (78/10)*(78/10)-(23/10)*(23/10)=(6084-529)/100=5555/100=1111/20