Explicație pas cu pas:
x²+4x+3=x²+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)
E(x) este definita pentru numitor diferit de 0 adica pt
(x+1)(x+3)=/=0
x+1=0=>x=-1
x+3=0=>x=-3
D=R\{-1,-3}
E(x)=(2x+1-x)(2x+1+x)/(x+1)(x+3)
Am folost a²-b²=(a+b)(a-b) pt (2x+1)²-x²
=>E(x)=(x+1)(3x+1)/(x+1)(x+3)=>
E(x)=(3x+1)/(x+3)
(3x+1)/(x+3)€Z <=> x+3|3x+1 dar x+3|x+3(il divide pe el insusi)
x+3|3x+1
x+3|x+3=>x+3|3x+9
Le scadem:
x+3|3x+9-3x-1=>x+3|8=>x+3€D8 in Z =>
x+3€{1,2,4,8,-1,-2,-4,-8} =>
x€{-2,-1,1,5,-4,-5,-7,-11}
Dar x€R\{-1,-3} => x€{-11,-7,-5,-4,-2,1,5}
