Determinăm domeniul de existență :
x+2 ≥ 0 ⇒ x ≥ -2 ⇒x ∈[-2, ∞)
x+3 ≥ 0 ⇒ x ≥ -3 ⇒x ∈[-3, ∞)
D = [-2, ∞) ∩ [-3, ∞) ⇒ D = [-2, ∞) (1)
Transformăm inecuația :
[tex]\it \sqrt{x+2} -\sqrt{x+3} \ \textless \ \sqrt2-\sqrt3 \Rightarrow \sqrt{x+2} +\sqrt3 \ \textless \ \sqrt{x+3} +\sqrt2\Rightarrow
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\Rightarrow (\sqrt{x+2} +\sqrt3)^2 \ \textless \ (\sqrt{x+3} +\sqrt2)^2\Rightarrow x+2+3+2\sqrt{3x+6} \ \textless \
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\ \textless \ x+3+2+2\sqrt{2x+6} \Rightarrow \sqrt{3x+6} \ \textless \ \sqrt{2x+6} \Rightarrow [/tex]
[tex]\it \Rightarrow 3x+6\ \textless \ 2x+6 \Rightarrow 3x\ \textless \ 2x\Rightarrow 3x-2x\ \textless \ 0\Rightarrow x\ \textless \ 0 \ \ \ \ \ (2)
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(1), \ (2) \Rightarrow x\in [-2,\ 0)[/tex]
Răspunsul corect este B).
