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a fost răspuns

Va rog ajutati-ma!Sa se rezolve inecuatiile:[tex] \frac{x+1}{x+2} \leq 2;

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Răspuns :

[tex] \frac{x+1}{x+2} \leq 2\\ \frac{x+1}{x+2}-2 \leq 0\\ \frac{x+1-2(x+2)}{x+2} \leq 0\\ \frac{x+1-2x-4}{x+2} \leq 0\\ \frac{-x-3}{x+2} \leq 0\\-x-3 \leq 0\\-x \leq 3\\x \leq -3\\ \\DVA\\x+2\ \textgreater \ 0\\x\ \textgreater \ -2\\[/tex]

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