Răspuns:
x∈{-2;-1;0;1}
Explicație pas cu pas:
x³-3x+2=x³-2x-x+2=x³-x-2x+2=x(x²-1)-2(x-1)=x(x-1) (x+1)-2(x-1)
=(x-1)(x²+x-2)=(x-1)(x²+2x-x-2)=(x-1) (x(x+2)-(x+2))=(x-1)(x+2) (x-1)=
(x-1)²(x+2)
(x-1)²(x+2)/(2x+1) ∈Z
a)(2x+1)|(x-1)
(2x+1)|(2x-2)
(2x+1)|(2x+1-3)
(2x+1)|3
2x+1∈D3Z={-3;-1;1;3}
2x∈{-4;-2;0;2}
x∈{-2;-1;0;1}
verificam ,(2x+1)|(x³-3x+2)
-2 convine, -1 convine, 0 convine, 1 convine
b)
(2x+1 )| (x+2)
(2x+1)|(2x+4)
(2x+1)|(2x+1+3)
(2x+1)|3
si avem tot x∈{-2;-1;0;1} deja verificate
deci
x∈{-2;-1;0;1}
