Question

Va rog ajutați-mă că nu stiu​

Answer 1

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}