Question

№ 2. Решите неравенства 1)√(x+2)>x 2) √(х^2+3)<2

Answer 1

$1) \ \sqrt{x+2}>x$

Неравенство вида $\sqrt{f(x)} > g(x)$ равносильно:

$\left[\begin{array}{ccc}\displaystyle \left \{ {{g(x) < 0, \ \ \ \ \ } \atop {f(x) \geq 0, \ \ \ \ \ }} \right. \\\displaystyle \left \{ {{g(x) \geq 0, \ \ \ \ } \atop {f(x) > g^{2}(x)}} \right.\\\end{array}\right$

Имеем:

$\left[\begin{array}{ccc}\displaystyle \left \{ {{x < 0, \ \ \ \ \ \ } \atop {x + 2 \geq 0, \ }} \right. \ \ \ \ (1) \\\displaystyle \left \{ {{x \geq 0, \ \ \ \ \ \ } \atop {x + 2 > x^{2}}} \right. \ \ \ \ \, (2) \\\end{array}\right$

$(1) \ \displaystyle \left \{ {{x < 0, \ \ \ \ \, } \atop {x + 2 \geq 0}} \right. \ \ \ \ \ \left \{ {{x < 0, \ \, } \atop {x \geq -2}} \right. \ \ \ \ \ \ x \in [-2; \ 0)$

$(2) \ \displaystyle \left \{ {{x \geq 0, \ \ \ \ \ \ } \atop {x + 2 > x^{2}}} \right. \ \ \ \ \ \left \{ {{x \geq 0, \ \ \ \ \ \ \ \ \ \ \ } \atop {x^{2} - x - 2 < 0}} \right. \ \ \ \ \ \ \left \{ {{x \geq 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \atop {(x+1)(x-2) < 0}} \right.$

$\displaystyle \left \{ {{x \geq 0, \ \ \ \ \ \ \ } \atop {-1<x<2}} \right. \ \ \ \ \ x \in [0; \ 2)$

Объединяем решения $(1)$ и $(2):$

$x \in [-2; \ 2)$

Ответ: $x \in [-2; \ 2)$

$2) \ \sqrt{x^{2} + 3} < 2$

Неравенство вида $\sqrt{f(x)} < a, \ a \geq 0,$ равносильно:

$\displaystyle \left \{ {{f(x) \geq 0,} \atop {f(x) < a^{2}}} \right.$

Имеем:

$1) \ x^{2} + 3 \geq 0; \ x^{2} \geq -3; \ x \in \mathbb{R}$

$2) \ x^{2} + 3 < 2^{2}$

$x^{2} < 1$

$\sqrt{x^{2}} < \sqrt{1}$

$|x| < 1$

$-1<x<1$

Ответ: $x \in (-1; \ 1)$