Răspuns:
Explicație pas cu pas:
[tex]E_{1}=\dfrac{2x-5}{3},~~ E_{2}=-\dfrac{4x+1}{5}.\\a)~E_{1}<E_{2},~~\dfrac{2x-5}{3}<-\dfrac{4x+1}{5}~|*15,~~5(2x-5)<-3(4x+1),~~10x-25<-12x-3,~~10x+12x<-3+25,~~22x<22~|:22,~~x<1.[/tex]
Deci x∈(-∞; 1).
[tex]b)~E_{1}+E_{2}\geq 0,~~\dfrac{2x-5}{3} +(-\dfrac{4x+1}{5} )\geq 0,~~\dfrac{2x-5}{3}- \dfrac{4x+1}{5}\geq0~|*15,~~5(2x-5)-3(4x+1)\geq 0,~~10x-25-12x-3\geq 0,~~-2x-28\geq 0,~~-2x\geq 28~|:(-2),~~x\leq -14[/tex]
Deci x∈(-∞; -14].
[tex]\left \{ {{E_{1}>0} \atop {E_{2}\leq 0}} \right.,~~\left \{ {{\frac{2x-5}{3}>0~|*3} \atop {-\frac{4x+1}{5} \leq 0~|*(-5)}} \right.,~~\left \{ {{2x-5>0} \atop {4x+1\geq 0}} \right.,~~\left \{ {{2x>5} \atop {4x\geq -1}} \right.,~~\left \{ {{x>\frac{5}{2} } \atop {x\geq -\frac{1}{4} }} \right. ,~~~deci~x>\frac{5}{2}[/tex]
x∈(5/2; +∞)
