Question

решить неравенство.......................................................................

Answer 1

$\sqrt{\dfrac{2+\sqrt{4-x^2}}{2}}+\sqrt{\dfrac{2-\sqrt{4-x^2}}{2}}\geq x$

ОДЗ неравенства $\displaystyle \left \{ {{4-x^2\geq 0} \atop {\dfrac{2-\sqrt{4-x^2}}{2}\geq 0}} \right. ~~~~\Rightarrow~~~-2\leq x\leq 2$

Возводим обе части неравенства до квадрата, получим

$\dfrac{2+\sqrt{4-x^2}}{2}+2\sqrt{\dfrac{2+\sqrt{4-x^2}}{2}\cdot \dfrac{2-\sqrt{4-x^2}}{2}}+\dfrac{2-\sqrt{4-x^2}}{2}\geq x^2\\ \\ \sqrt{4-4+x^2}\geq x^2-2\\ \\ |x|\geq x^2-2\\ \\ |x|^2-|x|-2\leq 0$

$|x|^2-|x|+\dfrac{1}{4}-\dfrac{9}{4}\leq 0\\ \\ \left(|x|-\dfrac{1}{2}\right)^2\leq \dfrac{9}{4}\\ \\ \left||x|-\dfrac{1}{2}\right|\leq \dfrac{3}{2}~~~\Rightarrow~~~ -\dfrac{3}{2}\leq |x|-\dfrac{1}{2}\leq \dfrac{3}{2}~~~\Rightarrow~~~-1\leq |x|\leq 2\\ \\ |x|\leq 2~~~\Rightarrow~~ -2\leq x \leq 2$

С учетом ОДЗ: x ∈ [-2;2].

Ответ: x ∈ [-2;2].

Answer 2

$\sqrt{\frac{2+\sqrt{4-x^2}}{2}}+\sqrt{\frac{2-\sqrt{4-x^2}}{2}}\geq x\; \; ,\\\\ODZ:\; \; \left\{\begin{array}{cc}4-x^2\geq 0\\\frac{2\pm \sqrt{4-x^2}}{2}\geq 0\end{array}\right\; \; \left\{\begin{array}{cc}-2\leq x\leq 2\qquad \\2\pm \sqrt{4-x^2}\geq 0\end{array}\right\; \; \left\{\begin{array}{cc}-2\leq x\leq 2\\\sqrt{4-x^2}\leq 2\end{array}\right\\\\\\\left\{\begin{array}{cc}-2\leq 2x\leq 2\\4-x^2\leq 4\end{array}\right\; \; \left\{\begin{array}{ccc}-2\leq x\leq 2\\x^2\geq 0\end{array}\right\; \; \Rightarrow \; \; \; -2\leq x\leq 2$

Воспользуемся формулой сложного корня:

$\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}\; ,$

все подкоренные выражения неотрицательны.

Если    $a=2\; ,\; b=x^2\; ,$  то  получим

$\sqrt{2+\sqrt{x^2}}\geq x\; \; \; \Leftrightarrow \; \; \left \{ {{x\geq 0} \atop {2+\sqrt{x^2}\geq x^2}} \right. \; \; ili\; \; \left \{ {{x\leq 0} \atop {2+\sqrt{x^2}\geq 0}} \right.\\\\1)\; \; x\geq 0\; \; \Rightarrow \; \; \sqrt{x^2}=|x|=x\; \; \; \; \Rightarrow \; \; \; 2+x\geq x^2\; ,\; \; x^2-x-2\leq 0\; ,\\\\x_1=-1\; ,\; x_2=2\; \; (teorema\; Vieta)\\\\(x+1)(x-2)\leq 0\; ,\qquad +++[-1\, ]---[\, 2\, ]+++\; \; \; x\in [-1,2\, ]\\\\x\geq 0\; \; \to \; \; x\in [\, 0,2\, ]$

$2)\; \; x\leq 0\; \; \to \; \; |x|=-x\; ,\\\\ 2+\sqrt{x^2}\geq 0\; \; ,\; \; 2+|x|\geq 0\; \; ,\; \; 2-x\geq 0\; \; ,\; \; x\leq 2\\\\x\leq 0\; \; \Rightarrow \; \; x\in (-\infty ,0\, ]\\\\3)\; \; \left\{ {{-2\leq x\leq 2} \atop {\left [ {{x\in [\, 0,2\, ]} \atop {x\in (-\infty ,0\, ]}} \right. }} \right. \; \; \Rightarrow \; \; \left \{ {{-2\leq x\leq 2} \atop {x\in [\, 0,2\, ]}} \right.\; \; ili\; \; \left \{ {{-2\leq x\leq 2} \atop {x\in (-\infty ,0\, ]}} \right. \; \; \; \Rightarrow$

$\left [ {{x\in [\, 0,2\, ]} \atop {x\in [-2,0\, ]}} \right. \; \; \; \Rightarrow \; \; \; x\in [-2,0\, ]\cup [\, 0,2\, ]=[-2,2\, ]\\\\Otvet:\; \; x\in [-2,2\, ]\; .$