45В. Решите неравенство  \(\frac{3}{{{{\left( {{2^{2-{x^2}}}-1} \right)}^2}}}-\frac{4}{{{2^{2-{x^2}}}-1}} + 1 \ge 0\).

Ответ

ОТВЕТ: \(\left( {-\infty ;\;-\sqrt 2 } \right) \cup \left( {-\sqrt 2 ;\;-1} \right] \cup \left\{ 0 \right\} \cup \left[ {1;\;\sqrt 2 } \right) \cup \left( {\sqrt 2 ;\;\infty } \right).\)

Решение

\(\frac{3}{{{{\left( {{2^{2-{x^2}}}-1} \right)}^2}}}-\frac{4}{{{2^{2-{x^2}}}-1}} + 1 \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{{{\left( {{2^{2-{x^2}}}-1} \right)}^2}-4 \cdot \left( {{2^{2-{x^2}}}-1} \right) + 3}}{{{{\left( {{2^{2-{x^2}}}-1} \right)}^2}}} \ge 0.\)

Пусть  \({2^{2-{x^2}}}-1 = t.\)  Тогда неравенство примет вид:  \(\frac{{{t^2}-4t + 3}}{{{t^2}}} \ge 0.\)

\({t^2}-4t + 3 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{t} = 3,}\\{{t} = 1.}\end{array}} \right.\)

\(\frac{{{t^2}-4t + 3}}{{{t^2}}} \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{\left( {t-1} \right)\left( {t-3} \right)}}{{{t^2}}} \ge 0.\)

Решим полученное неравенство методом интервалов:

\(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t \le 1,}\\{t \ne 0,}\end{array}} \right.}\\{t \ge 3}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{2^{2-{x^2}}}-1 \le 1,}\\{{2^{2-{x^2}}}-1 \ne 0,}\end{array}} \right.}\\{{2^{2-{x^2}}}-1 \ge 3}\end{array}} \right.\;\;\,\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{2^{2-{x^2}}} \le {2^1},\,}\\{{2^{2-{x^2}}} \ne {2^0},}\end{array}} \right.}\\{{2^{2-{x^2}}} \ge {2^2}}\end{array}} \right.\;\;\,\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{2-{x^2} \le 1,}\\{2-{x^2} \ne 0,}\end{array}} \right.}\\{2-{x^2} \ge 2}\end{array}} \right.\;\;\,\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{x^2}-1 \ge 0,\,}\\{2-{x^2} \ne 0,}\end{array}} \right.}\\{{x^2} \le 0\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\,\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{\left( {x-1} \right)\left( {x + 1} \right) \ge 0,}\\{{x^2} \ne 2,\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,\;}\end{array}} \right.}\\{{x^2} = 0\;\;\;\;\,\,\,\;\;\;\;\;\;\;\;\;\;\,\,}\end{array}} \right.\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \in \left( {-\infty ;-1} \right] \cup \left[ {1;\infty } \right),}\\{x \ne  \pm \sqrt 2 ,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,\;}\end{array}} \right.}\\{x = 0.\;\;\;\;\,\;\;\;\;\;\,\,\;\;\;\;\;\;\;\;\;\;\,\,}\end{array}} \right.\)

Таким образом, решением исходного неравенства является:

\(x \in \left( {-\infty ;\;-\sqrt 2 } \right) \cup \left( {-\sqrt 2 ;\;-1} \right] \cup \left\{ 0 \right\} \cup \left[ {1;\;\sqrt 2 } \right) \cup \left( {\sqrt 2 ;\;\infty } \right).\)

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