2) a+(a+2)+(a+4)+(a+6)=112
4a+12=112
4a=112-12
a=100/4
a=25
nr. impare consecutive sunt 25; 27; 29; 31
3) (3/4)·x>(1/3)·x+1,(6) 1,(6)=(16-1)/9=15/9 simplificam cu 3 =5/3
3x/4>x/3+5/3
3x/4-x/3>5/3 aducem la acelasi numitor
(9x-4x)/12>5/3
5x/12>5/3
5x·3>12·5
15x>60
x>60/15
x>4
X∈(4, +α)