Răspuns :

Triunghiul1id

-4/x+1 ∈ Z ⇒ x+1∈[tex]D_{4}[/tex];

x+1 ∈ {-1,1,-2,2,-4,4}/-1

x∈{-2,0,-3,1,-5,3}

______________________

-7/2x+1 ∈ Z ⇒ 2x+1 ∈ [tex]D_7[/tex]

2x+1 ∈ {-1,1,-7,7}/-1

2x ∈ {-2,0,-8,6}/:2

x ∈ {-1,0,-4,3}

______________________

2x+1/x+2 ∈ Z ⇒ x+2|2x+1

x+2|2x+1⇒x+2|2(x+2)

x+2|x+2⇒x+2|2x+4

x+2|2x+4-2x-1⇒x+2|3⇒x+2∈[tex]D_3[/tex]

x+2∈{-1,1,-3,3}/-2 ⇒ x∈{-3,-1,-5,1}
a. X+1| -4=> x+1€ Div lui -4 => x+1€ {1,-1,2,-2,-4,4} /-1 => x€{0,-2,1,-3,-5,3#
b. 2x+1| -7=> 2x+1€ div lui -7=> 2x+1€ {1,-1,7,-7}=> 2x€{0,-2,6,-8}=> x€{0,-1,3,-4}
c. x+2|2x+1 |. Din astea 2 afirmatii rezulta
x+2| x+2 | ca x+2 | divide si dif si suma lor
Inmultim la x+2|x+2 cu doi=> x+2| 2x+4=>
X+2|2x+1
X+2|2x+4 din acestea doua => ca x+2| 2x+1-(2x+4) =>. X+2 | 2x+1-2x-4 => x+2|-3=>x+2€ Div lui -3 => x+2=> {1,-1,3,-3} /-2=> x€{-1,-3,1,-5}