Question

Решите неравенство 96/(5^(2-x^2) - 1)^2-28/(5^(2-x^2)-1)+1>=0

Answer 1

$\dfrac{96}{(5^{2-x^2}-1)^2}-\dfrac{28}{5^{2-x^2}-1}+1 \geq 0 \\ \\3AMEHA:\ 5^{2-x^2}-1=t\\ \\ \dfrac{96}{t^2}-\dfrac{28}{t}+1 \geq 0\\ \dfrac{t^2-28t+96}{t^2} \geq 0\\ \dfrac{(t-4)(t-24)}{t^2} \geq 0$
  +       +        -        +
-----o-------|------|-------> t
      0        4       24
t < 0 или 0 < t ≤ 24 или t ≥ 24
$5^{2-x^2}-1\ \textless \ 0$ или $0\ \textless \ 5^{2-x^2}-1 \leq 4$ или $5^{2-x^2}-1 \geq 24$
$5^{2-x^2}\ \textless \ 5^0$ или $5^0\ \textless \ 5^{2-x^2} \leq 5^1$ или $5^{2-x^2} \geq 5^2$
$x^2-2\ \textgreater \ 0$ или $0\ \textless \ 2-x^2 \leq 1$ или $2-x^2 \geq 2$
х² > 2 или 1 ≤ х² < 2 или х² ≤ 0
$\left[ \begin{matrix} x \in (-\infty; - \sqrt{2}) \cup ( \sqrt{2};+\infty) \\ x \in (-\sqrt{2};-1] \cup [1; \sqrt{2}) \\ x=0 \end{matrix}\right$

Ответ: $(-\infty; - \sqrt{2}) \cup (-\sqrt{2};-1] \cup \{0\} \cup [1; \sqrt{2}) \cup ( \sqrt{2};+\infty).$