Răspuns:
Explicație pas cu pas:
b) E(x)=[4/(x-1)+(13-5x)/(1-x²)-(2x+10)/(x²+6x+5)]:(x-3)/(x²+3x+2)=
[-4/(1-x)+(13-5x)/(1-x²)-2(x+5)/(x+1)(x+5)]:(x-3)/(x+1)(x+2)=
[-4/(1-x)+(13-5x)/(1-x²)-2/(x+1)}·(x+1)(x+2)/(x-3)=
{[-4(x+1)+13-5x-2(1-x)]/(1-x)(1+x)}·(x+1)(x+2)/(x-3)=
[(-4x-4+13-5x-2+2x)/(1-x)(1+x)]·(x+1)(x+2)/(x-3)=
[(-7x+7)/(1-x)(1+x)]·(x+1)(x+2)/(x-3)=
[7(1-x)/(1-x)(1+x)]·(x+1)(x+2)/(x-3)=
[7/(1+x)]·(x+1)(x+2)/(x-3)=
7(x+1)(x+2)/(1+x)(x-3)=
7(x+2)/(x-3)
E(x)= 7(x+2)/(x-3)
c) E(x)∈Z ⇔ 7(x+2)/(x-3)∈Z ⇔ x-3=7 ⇔ x=10 ⇒ E(x)=7(10+2)/7=12∈Z
x-3=-7 ⇔ x=-4 ⇒ E(x)=7(-4+2)/-7=2∈Z
x-3=1 ⇔ x=4 ⇒ E(x)=7(4+2)/1=42∈Z
x-3=-1 ⇔ x=2 ⇒ E(x)=7(2+2)/-1=-28∈Z
