Răspuns:
Explicație pas cu pas:
Ex1. C, deoarece -2·x>-4 |:(-2), ⇒x<(-4):(-2), ⇒x<2.
Ex2. Produsul este negativ daca factorii au semne diferite, deci
[tex]\left \{ {{5x-2\geq 0} \atop {4x+3\leq0 }} \right. ~~sau~~\left \{ {{5x-2\leq 0} \atop {4x+3\geq 0 }} \right.~~\\1)~\left \{ {{5x-2\geq 0} \atop {4x+3\leq0 }} \right. ~~\left \{ {{5x\geq 2} \atop {4x\leq-3 }} \right. ~~\left \{ {{x\geq \frac{2}{5} } \atop {x\leq -\frac{3}{4} }} \right. ~n-are~solutii\\2)~\left \{ {{5x-2\leq 0} \atop {4x+3\geq 0 }} \right.~~\left \{ {{5x\leq 2} \atop {4x\geq -3 }} \right.~~\left \{ {{x\leq \frac{2}{5} } \atop {x\geq -\frac{3}{4} }} \right.[/tex]
Deci x∈[-3/4; 2/5].
Ex3.
[tex]\left \{ {{4x+1\leq 2x+7 } \atop {3x-20<8x-10}} \right. ~~\left \{ {{4x-2x\leq 7-1} \atop {3x-8x<-10+20}} \right. ~~\left \{ {{2x\leq 6} \atop {-5x<10}} \right. ~~\left \{ {{x\leq 3} \atop {x>-2}} \right.[/tex]
Deci x∈(-2; 3]
Ex4. Domeniul de definitie se afla din conditiile
[tex]\left \{ {{4x+16\geq 0} \atop {x+3\neq 0}} \right. ~~\left \{ {{4x\geq -16} \atop {x\neq -3}} \right. ~~\left \{ {{x\geq -4} \atop {x\neq -3}} \right.[/tex]
Deci Domeniul de definitie este x∈[-4; -3)∪(-3;+∞)
