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

[latex]\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[/latex]   +       +        -        + -----o-------|------|-------> t       0        4       24 t < 0 или 0 < t ≤ 24 или t ≥ 24 [latex]5^{2-x^2}-1\ \textless \ 0[/latex] или [latex]0\ \textless \ 5^{2-x^2}-1 \leq 4[/latex] или [latex]5^{2-x^2}-1 \geq 24[/latex] [latex]5^{2-x^2}\ \textless \ 5^0[/latex] или [latex]5^0\ \textless \ 5^{2-x^2} \leq 5^1[/latex] или [latex]5^{2-x^2} \geq 5^2[/latex] [latex]x^2-2\ \textgreater \ 0[/latex] или [latex]0\ \textless \ 2-x^2 \leq 1[/latex] или [latex]2-x^2 \geq 2[/latex] х² > 2 или 1 ≤ х² < 2 или х² ≤ 0 [latex]\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[/latex] Ответ: [latex] (-\infty; - \sqrt{2}) \cup (-\sqrt{2};-1] \cup \{0\} \cup [1; \sqrt{2}) \cup ( \sqrt{2};+\infty).[/latex]