Răspuns:
M1={-2/3}.
M2={0; 3}.
M3=∅
Explicație pas cu pas:
P1(x): |6x^2-5x-6| + |2x+3x^2| = 0 ⇔ |(3x+2)(2x-3)|+|x(3x+2)|=0;
|(3x+2)(2x-3)| ≥ 0, cu egalitate cand x=-2/3 sau x=3/2
|x(3x+2)| ≥ 0, cu egalitate cand x=0 sau x=-2/3
Asadar, |6x^2-5x-6| + |2x+3x^2| ≥ 0, cu egalitate cand x=-2/3.
In concluzie: M1={-2/3}.
P2(x): √(4x^2-12x+9) - 6|3-2x| + 15 = 0 ⇔ |2x-3| - 6|3-2x| + 15=0 ⇔
⇔ -5|2x-3| = -15 ⇔ |2x-3| = 3 ⇔ 2x-3=3 (⇔x=3) sau 2x-3=-3 (⇔x=0).
In concluzie: M2={0; 3}.
P3(x): [x-1/3]=x+1/4; Cum [x-1/3] ∈ Z, x+1/4 ∈ Z, deci {x} = 1-1/4 = 3/4.
Insa [x-1/3]=x-1/3-{x-1/3}=x-3/4
Obtinem: x-3/4=x+1/4 ⇒ x ∈ ∅
