Question

Помогите решить неравенство

Answer 1

-23 / ((x + 3)² - 6) ≥ 0 |:-23

1 /  ((x + 3)² - 6) ≤ 0

a² - b² = (a - b)(a + b)

1 /  ((x + 3 + √6)(x + 3 - √6)) ≤ 0

+++++++(-3-√6) ---------------- (-3+√6) +++++++++

x ∈ (-3-√6, -3+√6)

Answer 2

$\frac{-23}{(x+3)^{2}-6 }\geq0 \ \Rightarrow (x+3)^{2}-6<0 \\\\x^{2}+6x+9-6<0\\\\x^{2}+6x+3<0\\\\x^{2}+6x+3=0\\\\D=6^{2}-4*3=36-12=24=(2\sqrt{6})^{2}\\\\x_{1} =\frac{-6-2\sqrt{6}}{2}=-3-\sqrt{6}\\\\x_{2}=\frac{-6+2\sqrt{6}}{2}=\sqrt{6} -3 \\\\x^{2} +6x+3=[x-(-3-\sqrt{6})][x-(\sqrt{6} -3)]$

$[x-(-3-\sqrt{6})][x-(\sqrt{6} -3)]<0$

    +               -                       +

______₀__________₀________

        - 3 - √6            √6 - 3                    

           /////////////////////////

Ответ : x ∈ (- 3 - √6 ; √6 - 3)