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Stie cineva? :)
Determinati x e R pentru care 0 ≤ (x+1)/2 ≤ 3


Răspuns :

[tex]0\leq \dfrac{x+1}{2}\leq3|_{\cdot 2} \Longleftrightarrow 0\leq x+1\leq6|_{-1}\Longleftrightarrow -1\leq x \leq 5 \\\;\\ x\in [-1,\ 5][/tex]