Условие задачи
Анна Малкова Решите неравенство:
\(\displaystyle 2^{x-2}\cdot (2^x-3) \leq \left (\frac{1}{4-2^x}-1 \right )\)
Решение
Замена \(2^x = t, \, \, t \, \textgreater \, 0\)
\(\displaystyle t^2-3t \leq 4 (\frac{1}{4-t}-1)\)
\(\displaystyle t^2-3t+4 \leq \frac{4}{4-t}\)
\(\displaystyle t^2-3t+4+\frac{4}{t-4} \leq 0\)
\(\displaystyle t^2-3t+4(1+\frac{1}{t-4}) \leq 0\)
\(\displaystyle t(t-3)+4 \cdot \frac{t-3}{t-4} \leq 0\)
\(\displaystyle (t-3)(t+ \frac{4}{t-4}) \leq 0\)
\(\displaystyle (t-3)(\frac{t^2-4t+4}{t-4}) \leq 0\)
\(\displaystyle \frac{(t-3)(t-2)^2}{t-4} \leq 0\)
Решим неравенство методом интервалов.
\(\left[ \begin{gathered} t=2 \\ 3 \leq t \, \textless \, 4 \\ \end{gathered} \right. ; \, \, \left[ \begin{gathered} 2^x=2 \\ 3 \leq 2^x\, \textless \, 4 \\ \end{gathered} \right. \, \, ;\left[ \begin{gathered} x=1 \\ log_2 3 \leq x \, \textless \, 2\\ \end{gathered} \right.\)
Ответ
\(\left \{ 1 \right \} \cup [log_2 3; 2)\)
