Question

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|x+2| * log1/2 din (2-x) >= 0

Answer 1

$|x+2|\cdot \log_{\frac{1}{2}}(2-x) \geq 0\Leftrightarrow \\ \\ |x+2| > 0,\quad \forall x\in \mathbb{R} \Rightarrow x = -2\quad (1) \\ \\ (2)\quad \log_{\frac{1}{2}}(2-x) \geq 0\,\Bigg|(\frac{1}{2})^{()}\\\\ a^x\text{ functie strict descrescatoare cand } 0<a<1 \\ \Rightarrow x_1 \geq x_2 \Leftrightarrow a^{x_1}\leq a^{x_2}\\ \\ \\\Leftrightarrow (\frac{1}{2})^{ \log_{\frac{1}{2}}(2-x)} \leq (\frac{1}{2})^{0}$

$\\ \\ \Leftrightarrow 2-x \leq 1 \Leftrightarrow 1\leq x \Leftrightarrow x\geq 1,\quad \text{dar din c.d. }2-x> 0 \Leftrightarrow x < 2 \\ \\ \\\Rightarrow x\in (1)\cup (2) \Leftrightarrow \\ \Leftrightarrow x\in \{-2\}\cup \Big([1,+\infty)\cap (-\infty, 2)\Big) \Rightarrow \boxed{x\in \{-2\}\cup\big[1,2\big)}$

Answer 2

Răspuns:

Explicație pas cu pas:

|x+2| * log1/2 din (2-x) >= 0