Question

$\frac{x + 2}{3 - x} > (0.5) {}^{ - 1}$
решите неравенство ​

Answer 1

$\frac{x+2}{3-x}>(0,5)^{-1}\\\\\frac{x+2}{3-x}>(\frac{1}{2} )^{-1}\\\\\frac{x+2}{3-x}>2\\\\\frac{x+2}{3-x}-2>0\\\\\frac{x+2-6+2x}{3-x}>0\\\\\frac{3x-4}{3-x}>0\\\\\frac{3(x-1\frac{1}{3}) }{-(x-3)}>0\\\\\frac{x-1\frac{1}{3} }{x-3}<0$

     +              -              +

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          1 1/3            3

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Ответ : x ∈ (1 1/3  ;  3)