Question

решите АЖПЖПЖПЖПЖ8 < x ^ 2 - 4x + 3 <= 15​

Answer 1

Объяснение:

$\displaystyle\\8 < x^2-4x+3\leq 15\ \ \ \ \ \ \Rightarrow\\\\\left \{ {{x^2-4x+3 > 8} \atop {x^2-4x+3\leq 15}} \right. \ \ \ \ \ \ \ \left \{ {{x^2-4x-5 > 0} \atop {x^2-4x-12\leq 0}} \right.$

$\displaystyle\\1.\ \\x^2-4x-5 > 0\\\\x^2-5x+x-5 > 0\\\\x*(x-5)+(x-5) > 0\\\\(x-5)*(x+1) > 0$

-∞__+__-1__-__5__+__+∞

x∈(-∞;-1)U(5;+∞).

$\displaystyle\\2.\\x^2-4x-12\leq 0\\\\x^2-6x+2x-12\leq 0\\\\x*(x-6)+2*(x-6)\leq 0\\\\(x-6)*(x+2)\leq 0$

-∞__+__-2__-__6__+__+∞

x∈[-2;6].                 ⇒

                                      Ответ: x∈[-2;-1)U(5;6].