Đáp án:
$24\, B. \, \dfrac{1}{e^2} < x < e$
$25\, C. \, S = \dfrac32$
Giải thích các bước giải:
$\begin{array}{l}24)\quad \dfrac{\ln x + 2}{\ln x - 1} < 0 \qquad (x >0)\\ \Leftrightarrow \left[\begin{array}{l}\begin{cases}\ln x + 2 > 0\\\ln x - 1 < 0\end{cases}\\ \begin{cases}\ln x + 2 < 0\\\ln x - 1 > 0\end{cases}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\begin{cases}\ln x >-2\\\ln x <1\end{cases}\\ \begin{cases}\ln x <-2\\\ln x >1\end{cases}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\begin{cases}x > e^{-2}\\x < e\end{cases}\\ \begin{cases}x < e^{-2}\\x > e\end{cases}\quad (loại)\end{array}\right.\\ \Leftrightarrow \dfrac{1}{e^2} < x < e\\ 25)\quad 81^x - 4.3^{2x+1} + 27 = 0\\ \Leftrightarrow 3^{4x} - 4.3^{2x}.3 + 27 = 0\\ \Leftrightarrow \left(3^{2x}\right)^2 - 12.3^{2x} + 27 = 0\\ \Leftrightarrow \left[\begin{array}{l}3^{2x} =9\\3^{2x} = 3 \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = 2\\2x = 1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 1\\x = \dfrac12\end{array}\right.\\ \Rightarrow S = 1 + \dfrac12 = \dfrac32 \end{array}$
