35В. Решите неравенство \({2^{2x-{x^2}-1}} + \frac{1}{{{2^{2x-{x^2}}}-1}} \le 2\).
\({2^{2x-{x^2}-1}} + \frac{1}{{{2^{2x-{x^2}}}-1}} \le 2\;\;\;\; \Leftrightarrow \;\;\;\frac{{{2^{2x-{x^2}}}}}{2} + \frac{1}{{{2^{2x-{x^2}}}-1}} \le 2.\) Пусть \({2^{2x-{x^2}}} = t.\) Тогда неравенство примет вид: \(\frac{t}{2} + \frac{1}{{t-1}}-2 \le 0\left| { \cdot 2} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{{t^2}-t + 2-4t + 4}}{{t-1}} \le 0\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{{t^2}-5t + 6}}{{t-1}} \le 0.\) \({t^2}-5t + 6 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{t} = 2,}\\{{t} = 3.}\end{array}} \right.\) \(\frac{{{t^2}-5t + 6}}{{t-1}} \le 0\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{\left( {t-2} \right)\left( {t-3} \right)}}{{t-1}} \le 0.\) Решим полученное неравенство методом интервалов: \(\left[ {\begin{array}{*{20}{c}}{t < 1,\;\;\;\;\;}\\{2 \le t \le 3}\end{array}} \right.\;\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{2^{2x-{x^2}}} < {2^0},\;\;\;\;\;\;\;\;\;}\\{{2^1} \le {2^{2x-{x^2}}} \le {2^{{{\log }_2}3}}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{2x-{x^2} < 0,\;\;\;\;\;\;\;\;\;\;\,}\\{1 \le 2x-{x^2} \le {{\log }_2}3}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2}-2x > 0,\,\,\;\;\;\;\;\;\;\;\,}\\{\left\{ {\begin{array}{*{20}{c}}{2x-{x^2} \ge 1,\;\;\;\;\;}\\{2x-{x^2} \le {{\log }_2}3}\end{array}\;\;\,} \right.}\end{array}} \right. \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x\left( {x-2} \right) > 0,\;\,\,\;\;\;\;\;\;\;\;\,}\\{\left\{ {\begin{array}{*{20}{c}}{{x^2}-2x + 1 \le 0,\,\;\;\;\;\;}\\{{x^2}-2x + {{\log }_2}3 \ge 0}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x\left( {x-2} \right) > 0,\;\,\,\;\;\;\;\;\;\;\;\,}\\{\left\{ {\begin{array}{*{20}{c}}{{{\left( {x-1} \right)}^2} \le 0,\,\;\;\;\;\;\;\;\,\;\;}\\{{x^2}-2x + {{\log }_2}3 \ge 0}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x \in \left( {-\infty ;0} \right) \cup \left( {2;\infty } \right),}\\{\left\{ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-2x + {{\log }_2}3 \ge 0.}\end{array}\;\;} \right.}\end{array}} \right.\) Решим неравенство: \({x^2}-2x + {\log _2}3 \ge 0;\,\,\,\,\, D = 4-4{\log _2}3 < 0;\,\,\,\,\,\,\,\,x \in R.\) Следовательно: \(\left[ {\begin{array}{*{20}{c}}{x \in \left( {-\infty ;0} \right) \cup \left( {2;\infty } \right),}\\{\left\{ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \in R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\;\;} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;0} \right) \cup \left\{ 1 \right\} \cup \left( {2;\infty } \right).\) Ответ: \(\left( {-\infty ;0} \right) \cup \left\{ 1 \right\} \cup \left( {2;\infty } \right).\)