95В (ЕГЭ 2025). Решите неравенство \(\dfrac{{105}}{{{{\left( {{2^{4-{x^2}}}-1} \right)}^2}}}-\dfrac{{22}}{{{2^{4-{x^2}}}-1}} + 1 \ge 0\).
ОТВЕТ: \(\left( {-\infty ;\;-2} \right) \cup \left( {-2;\;-1} \right] \cup \left\{ 0 \right\} \cup \left[ {1;\,2} \right) \cup \left( {2;\;\infty } \right).\)
\(\dfrac{{105}}{{{{\left( {{2^{4-{x^2}}}-1} \right)}^2}}}-\dfrac{{22}}{{{2^{4-{x^2}}}-1}} + 1 \ge 0\) Пусть \({2^{4-{x^2}}}-1 = t.\) Тогда неравенство примет вид: \(\dfrac{{105}}{{{t^2}}}-\dfrac{{22}}{t} + 1 \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{{t^2}-22t + 105}}{{{t^2}}} \ge 0.\) \({t^2}-22t + 105 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{t = 7,}\\{t = 15.}\end{array}} \right.\) \(\dfrac{{{t^2}-22t + 105}}{{{t^2}}} \ge 0\;\;\;\; \Leftrightarrow \,\;\;\;\dfrac{{\left( {t-7} \right)\left( {t-15} \right)}}{{{t^2}}} \ge 0.\) Решим полученное неравенство методом интервалов: \(\left[ {\begin{array}{*{20}{c}}{\left\{ \begin{array}{l}t \le 7,\\t \ne 0\end{array} \right.}\\{t \ge 15}\end{array}} \right.\;\;\;\; \Leftrightarrow \,\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ \begin{array}{l}{2^{4-{x^2}}}-1 \le 7,\\{2^{4-{x^2}}}-1 \ne 0,\end{array} \right.}\\{{2^{4-{x^2}}}-1 \ge 15}\end{array}} \right.\;\;\;\; \Leftrightarrow \,\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ \begin{array}{l}{2^{4-{x^2}}} \le {2^3},\\{2^{4-{x^2}}} \ne {2^0},\end{array} \right.}\\{{2^{4-{x^2}}} \ge {2^4}}\end{array}} \right.\;\;\;\; \Leftrightarrow \,\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ \begin{array}{l}4-{x^2} \le 3,\\4-{x^2} \ne 0,\end{array} \right.}\\{4-{x^2} \ge 4}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ \begin{array}{l}{x^2} \ge 1,\\{x^2} \ne 4,\end{array} \right.}\\{{x^2} \le 0\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ \begin{array}{l}x \in \left( {-\infty ;-1} \right] \cup \left[ {1; + \infty } \right),\\x \ne \pm 2,\end{array} \right.}\\{x = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \in \left( {-\infty ;-2} \right) \cup \left( {-2;-1} \right] \cup \left\{ 0 \right\} \cup \left[ {1;2} \right) \cup \left( {2; + \infty } \right).\) Таким образом, решением исходного неравенства является: \(x \in \left( {-\infty ;\;-2} \right) \cup \left( {-2;\;-1} \right] \cup \left\{ 0 \right\} \cup \left[ {1;\,2} \right) \cup \left( {2;\;\infty } \right).\) Ответ: \(\left( {-\infty ;\;-2} \right) \cup \left( {-2;\;-1} \right] \cup \left\{ 0 \right\} \cup \left[ {1;\,2} \right) \cup \left( {2;\;\infty } \right).\) 