Răspuns :

Miky93id
aducem la acelasi numitor

18 < 4x+8 < 27 |-8

10 < 4x < 19 |:4

2,5 < x < 4,75

x∈N  ⇒x∈(2;4)  ⇔x∈{3}
[tex]\dfrac32<x+\dfrac23<\dfrac94\ scadem \ \dfrac23\Rightarrow\dfrac32-\dfrac23<x<\dfrac94-\dfrac23\Rightarrow\dfrac56<x<\dfrac{19}{12}\Rightarrow [/tex]

[tex]\Rightarrow0,...<x<1,...\Rightarrow x=1[/tex]

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