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Seby233
a fost răspuns

1. Determinati valoarea lui ,,A'' din :
a) (12+a)+14=4xa-4
b) (3xa+2):4-2=3
c) [(a+3)x3+3]x3+3=102
d) 5xa:2=2525

2. Ordonati crescator numerele :
A= 7360 - [(5+206x8)-709]
B= 6+8X[14X6+3X(7X9-4X5]
C= 2+10X[632+10X(14+14:7)]
D= 40+3X{32:8+3X[50+3X(200:4-98:2)]}


Răspuns :

1
a/ 12+a+14=4xa-4 ⇒ 12+14+4=4xa-a ⇒ 30= a(4x-1) ⇒ a= 30/4x-1
b/ 3xa+2-8=12 ⇒ 3xa= 18 ⇒ a= 18/3x= 6/x
c/ (3a+9+3)*3+3=102 ⇒ 9a+27+9+3= 102 ⇒9a + 39= 102 ⇒ 9a = 63 ⇒ a=63/9=7
d/ 5xa=2525*2 ⇒ 5xa=5050 ⇒ a= 5050/5x ⇒ a= 1010/x
2
A= 7360-(5+1648)-709= 7360 - 1653-709= 4998
B= 6+8·[(84+3·(63-20)]= 6+8·(84+129)= 6+1704= 1710
C=2+10·(632+160)= 2+7920= 7922
D=40+3·[4+3·(50+3·1)= 40+3·(4+159)=40+489=529
D,B,A,C respectiv 529,1710,5998,7922