Răspuns:
Explicație pas cu pas:
[tex]E(x)=\frac{(2x+1)^2-x^{2} }{x^{2} +4x+3} =\frac{(2x+1+x)(2x+1-x)}{x^{2} +4x+3} =\frac{(3x+1)(x+1)}{x^{2} +4x+3}[/tex]
Domeniul de existenta
x²+4x+3≠0
(x+1)(x+3)≠0
x≠-1
x≠-3
x∈R\{-3,-1}
[tex]E(x) =\frac{(3x+1)(x+1)}{x^{2} +4x+3}=\frac{(3x+1)(x+1)}{(x+1)(x+3)}=\frac{3x+1}{x+3}[/tex]
3x+1 | x+3 |×3
3x+1 |3x+9
3x+1 | 3x+1
facem diferenta
3x+1 | 8
3x+1={1,-1,2,-2,4,-4,8,-8}
3x={0,-2,1,-3,3,-5,7,-9}
x∈Z
x={0,-1,1,-3}
dar x nu are voie sa fie -1 si -3
deci x={0,1}
