Question

Sa se determine multimea de adevar a predicatelor:
p1(x): |6x^2-5x-6| + |2x+3x^2| = 0
p2(x): radical din 4x^2-12x+9 (radical inchis) - 6|3-2x| + 15 = 0
p3(x): [x-1/3] = x+1/4

Answer 1

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 ∈ ∅