Răspuns:
[tex]\bf x\in \Big[3,\dfrac{11}{3}\Big)[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf \Big[\dfrac{3x+1}{2} \Big]=5[/tex]
[tex]\bf 5\leq \dfrac{3x+1}{2} <6\:\:\Big|\cdot2\:\:\:\:\text{\it(Inmultim toata relatia cu 2)}[/tex]
[tex]\bf 5\cdot 2\leq 3x+1 <6\cdot2[/tex]
[tex]\bf 10 \leq 3x+1 <12\:\:\:\Big-1\:\:\:\:\text{\it(Scadem toata relatia cu 1)}[/tex]
[tex]\bf 10-1 \leq 3x+1-1 <12-1[/tex]
[tex]\bf 9 \leq 3x <11\:\:\Big|:3\:\:\:\:\text{\it(Inpartim toata relatia cu 3)}[/tex]
[tex]\bf 3 \leq 3x <\dfrac{11}{3}[/tex]
[tex]\boxed{\bf x\in \Big[3,\dfrac{11}{3}\Big)}[/tex]
==pav38==
