a) Điều kiện: \(\displaystyle \left\{ \begin{array}{l}4x + 2 > 0\\x - 1 > 0\\x > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > - \frac{1}{2}\\x > 1\\x > 0\end{array} \right. \Leftrightarrow x > 1\).
Khi đó \(\displaystyle \ln (4x + 2) - \ln (x - 1) = \ln x\)\(\displaystyle \Leftrightarrow \ln (4x + 2) = \ln [x(x - 1){\rm{]}}\)
\(\displaystyle \Leftrightarrow 4x + 2 = {x^2} - x\) \(\displaystyle \Leftrightarrow {x^2} - 5x - 2 = 0\)
\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5 + \sqrt {33} }}{2}\\x = \frac{{5 - \sqrt {33} }}{2}(l)\end{array} \right.\)\(\displaystyle \Leftrightarrow x = \frac{{5 + \sqrt {33} }}{2}\)
Vậy phương trình có nghiệm \(\displaystyle x = \frac{{5 + \sqrt {33} }}{2}\).
b) ĐK: \(\displaystyle \left\{ \begin{array}{l}3x + 1 > 0\\x > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > - \frac{1}{3}\\x > 0\end{array} \right. \Leftrightarrow x > 0\).
Khi đó:
\(\displaystyle {\log _2}(3x + 1){\log _3}x = 2{\log _2}(3x + 1)\) \(\displaystyle \Leftrightarrow {\log _2}(3x + 1){\rm{[}}{\log _3}x - 2] = 0\)
\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}{\log _2}(3x + 1) = 0\\{\log _3}x - 2 = 0\end{array} \right.\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}3x + 1 = 1\\{\log _3}x = 2\end{array} \right.\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x = 0(l)\\x = 9\end{array} \right. \Leftrightarrow x = 9\).
c) ĐK: \(\displaystyle x > 0\). Khi đó,
\(\displaystyle {2^{{{\log }_3}{x^2}}}{.5^{{{\log }_3}x}} = 400\)\(\displaystyle \Leftrightarrow {4^{{{\log }_3}x}}{.5^{{{\log }_3}x}} = 400\) \(\displaystyle \Leftrightarrow {20^{{{\log }_3}x}} = {20^2}\) \(\displaystyle \Leftrightarrow {\log _3}x = 2 \Leftrightarrow x = 9\) (TM)
d) ĐK: \(\displaystyle x > 0\).
Đặt \(\displaystyle t = \ln x\), ta có phương trình:
\(\displaystyle {t^3} - 3{t^2} - 4t + 12 = 0\)\(\displaystyle \Leftrightarrow \left( {t - 2} \right)\left( {t + 2} \right)\left( {t - 3} \right) = 0\)\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t = 2\\t = - 2\\t = 3\end{array} \right.\)
\(\displaystyle \Rightarrow \left[ \begin{array}{l}\ln x = 2\\\ln x = - 2\\\ln x = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = {e^2}\\x = {e^{ - 2}}\\x = {e^3}\end{array} \right.\)
