Неравенство

[Гост]

Моля за помощ с това нeрaвeнcтвo

[tex]\frac{1}{\sqrt{2-x}}≥\frac{1}{1+x}[/tex]

[ammornil]

Моля за помощ с това нeрaвeнcтвo $$\frac{1}{\sqrt{2-x}}≥\frac{1}{1+x}$$

Може би нещо такова...

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[tex]\text{ДМ: }\hspace{0.4em} \begin{array}{|l} 2-x>0 \\ 1+x\ne 0 \end{array} \Leftrightarrow \hspace{0.4em} \begin{array}{|l} x<2 \\ x\ne -1 \end{array} \Leftrightarrow x\in (-\infty; -1) \cup (-1,2)[/tex]

[tex](1) \hspace{0.4em} x\in (-\infty; -1) \rightarrow \frac{1}{\sqrt{2-x}}≥\frac{1}{1+x}, \hspace{0.4em} \forall x\in (-\infty; -1)[/tex]
[tex](2) \hspace{0.4em} x \in (-1; 2) \rightarrow \frac{1}{\sqrt{2-x}}≥\frac{1}{1+x} \hspace{0.4em}|(\cdots)^{2} \Leftrightarrow \hspace{0.4em} \frac{1}{2-x}\ge \frac{1}{x^{2}+2x+1} \Leftrightarrow \hspace{0.4em} x^{2}+2x+1 \ge 2-x \Leftrightarrow \hspace{0.4em} x^{2}+3x-1\ge 0[/tex]
[tex]\hspace{2em} x_{1,2}=\frac{-3\pm \sqrt{3^{2}-4\cdot{1}\cdot{-1}}}{2\cdot{1}}=\frac{-3\pm \sqrt{13}}{2} \hspace{0.4em} \Rightarrow \begin{array}{|l} x \in (-1; 2) \\ x \in \left(-\infty; \underbrace{\frac{-3-\sqrt{13}}{2}}_{\approx{-3.3} < -1}\right] \cup \left[\underbrace{\frac{-3+\sqrt{13}}{2}}_{-1<\approx{0,3}<2}; +\infty\right) \end{array} \Rightarrow x \in \left[\frac{-3+\sqrt{13}}{2}; +\infty\right)[/tex]
$$ x \in (-\infty; -1) \cup \left[\frac{-3+\sqrt{13}}{2}; +\infty\right)$$