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[tex]Rezolvati~inecuatiile~in~R: \\ \frac{x \sqrt{3}+1 }{3 x^{2}+2x \sqrt{3} +3 } \leq 0[/tex]

Răspuns :

[tex]\frac{x\sqrt3+1}{3x^2+2x\sqrt3+3}\leq 0\\ Notam:x\sqrt3+1=a\\ Si\ inlocuim:\\ \frac{a}{a^2+2} \leq 0\\ Ne\ putem\ da\ usor\ seama\ ca\ a^2+2 \geq 0 \Rightarrow a\leq 0\\ Inlocuind:\\ x\sqrt3+1 \leq 0\\ x\sqrt3 \leq-1\\ x\leq -\frac{\sqrt3}{3}\\ S:x \in (-infinit,-\frac{\sqrt3}{3}][/tex]