61В. Решите неравенство  \(\frac{{{{27}^{x + \frac{1}{3}}}-10 \cdot {9^x} + 10 \cdot {3^x}-5}}{{{9^{x + \frac{1}{2}}}-10 \cdot {3^x} + 3}} \le {3^x} + \frac{1}{{{3^x}-2}} + \frac{1}{{{3^{x + 1}}-1}}\).

Ответ

ОТВЕТ: \(\left( {-\infty ;\;-1} \right) \cup \left( {-1;\;0} \right] \cup \left( {{{\log }_3}2;\;1} \right).\)

Решение

\(\frac{{{{27}^{x + \frac{1}{3}}}-10 \cdot {9^x} + 10 \cdot {3^x}-5}}{{{9^{x + \frac{1}{2}}}-10 \cdot {3^x} + 3}} \le {3^x} + \frac{1}{{{3^x}-2}} + \frac{1}{{{3^{x + 1}}-1}}\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{3 \cdot {3^{3x}}-10 \cdot {3^{2x}} + 10 \cdot {3^x}-5}}{{3 \cdot {3^{2x}}-10 \cdot {3^x} + 3}} \le {3^x} + \frac{1}{{{3^x}-2}} + \frac{1}{{3 \cdot {3^x}-1}}.\)

Пусть  \({3^x} = t.\)  Тогда неравенство примет вид:

\(\frac{{3{t^3}-10{t^2} + 10t-5}}{{3{t^2}-10t + 3}}-t \le \frac{1}{{t-2}} + \frac{1}{{3t-1}}\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\frac{{3{t^3}-10{t^2} + 10t-5-3{t^3} + 10{t^2}-3t}}{{3{t^2}-10t + 3}} \le \frac{{4t-3}}{{\left( {3t-1} \right)\left( {t-2} \right)}}\,\,\,\,\,\, \Leftrightarrow \;\;\;\;\frac{{7t-5}}{{3{t^2}-10t + 3}} \le \frac{{4t-3}}{{\left( {3t-1} \right)\left( {t-2} \right)}}.\)

\(3{t^2}-10t + 3 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{t} = \frac{1}{3},}\\{{t} = 3.}\end{array}} \right.\)

\(3{t^2}-10t + 3 = \left( {3t-1} \right)\left( {t-3} \right).\)

\(\frac{{7t-5}}{{3{t^2}-10t + 3}} \le \frac{{4t-3}}{{\left( {3t-1} \right)\left( {t-2} \right)}}\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{7t-5}}{{\left( {3t-1} \right)\left( {t-3} \right)}}-\frac{{4t-3}}{{\left( {3t-1} \right)\left( {t-2} \right)}} \le 0\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\frac{{7{t^2}-14t-5t + 10-4{t^2} + 12t + 3t-9}}{{\left( {3t-1} \right)\left( {t-3} \right)\left( {t-2} \right)}} \le 0\;\;\;\; \Leftrightarrow \;\;\;\;\frac{{3{t^2}-4t + 1}}{{\left( {3t-1} \right)\left( {t-3} \right)\left( {t-2} \right)}} \le 0.\)

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

\(3{t^2}-4t + 1 = \left( {3t-1} \right)\left( {t-1} \right).\)

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

Решим полученное неравенство методом интервалов:

\(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t \ne \frac{1}{3},}\\{t \le 1,}\end{array}\,\,\,} \right.}\\{2 < t < 3}\end{array}} \right.\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{3^x} \ne \frac{1}{3},}\\{{3^x} \le 1,}\end{array}\,\,\,} \right.}\\{2 < {3^x} < 3}\end{array}} \right.\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{3^x} \ne {3^{-1}},\;\;\;\;\,\;\;}\\{{3^x} \le {3^0},\;\;\;\;\;\;\;\;\,}\end{array}} \right.}\\{{3^{{{\log }_3}2}} < {3^x} < {3^1}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \ne -1,\,\,\,\,\,\,}\\{x \le 0,\;\,\;\,\;\;}\end{array}} \right.\,\,\,\,\,\,}\\{{{\log }_3}2 < x < 1.}\end{array}} \right.\)

Таким образом, решением исходного неравенства является:  \(x \in \left( {-\infty ;\;-1} \right) \cup \left( {-1;\;0} \right] \cup \left( {{{\log }_3}2;\;1} \right).\)

Ответ:  \(\left( {-\infty ;\;-1} \right) \cup \left( {-1;\;0} \right] \cup \left( {{{\log }_3}2;\;1} \right).\)