Question

Determinați elementele mulțimilor (Dau 30 puncte)
A={x| x ∈ Z, 7(2x+1) ∈ Z}
B={x| x ∈ Z, $\frac{3x-2}{x-1}$ ∈ Z}
C={x| $\frac{2x+1}{3x+1}$∈ Z}
D={x| x∈ Z,$\frac{x^{2} +7x-8}{x^{2}-1 }$∈ Z}

Answer 1

Explicație pas cu pas:

(3x-2)/(x-1)€Z <=>x-1 | 3x-2 dar stim ca

x-1|x-1 deci avem:

x-1|x-1/×3=>x-1|3x-3

x-1|3x-2

Le scadem si ramanem cu:

x-1|3x-3-3x+2 => x-1|-1 => x-1€D -1 inZ =>

x-1€{-1,1} => x€{0,2}=> B€{0,2}

(2x+1)/(3x+1)€Z <=>3x+1|2x+1 deci avem:

3x+1|3x+1/×2=>3x+1|6x+2

3x+1|2x+1/×3=>3x+1|6x+3

Le scadem si avem:

3x+1|6x+2-6x-3=> 3x+1|-1 => 3x+1€D-1 inZ => 3x+1€{-1,1} => 3x€{-2,0} => x€{-2/3,0} deci C={0}

(x²+7x-8)/(x²-1)=(x²+8x-x-8)/(x+1)(x-1)=

(x+8)(x-1)/(x+1)(x-1)=(x+8)/(x+1)

(x+8)/(x+1)€Z <=>x+1|x+8 deci :

x+1|x+1

x+1|x+8

Le scadem si avem:

x+1|x+1-x-8 => x+1|-7 => x+1€D -7 in Z =>

x+1€{-7,-1,1,7} => x€{-8,-2,0,6} =>

D€{-8,-2,0,6}