Bài tập 2.68 trang 133 SBT Toán 12

a) ĐK: \(\left\{ \begin{array}{l}
4x + 2 > 0\\
x - 1 > 0\\
x > 0
\end{array} \right. \Leftrightarrow x > 1\)

\(\begin{array}{l}
\ln \left( {4x + 2} \right) - \ln \left( {x - 1} \right) = \ln x\\
 \Leftrightarrow \ln \left( {4x + 2} \right) = \ln x + \ln \left( {x - 1} \right)\\
 \Rightarrow 4x + 2 = x\left( {x - 1} \right)\\
 \Leftrightarrow {x^2} - 5x - 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{5 + \sqrt {33} }}{2}\\
x = \frac{{5 - \sqrt {33} }}{2}\,\,\left( l \right)
\end{array} \right.
\end{array}\)

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:

\(\displaystyle {\log _2}(3x + 1){\log _3}x = 2{\log _2}(3x + 1)\)

\(\Leftrightarrow {\log _2}\left( {3x + 1} \right).{\log _3}x - 2{\log _2}\left( {3x + 1} \right) = 0\)

\(\displaystyle  \Leftrightarrow {\log _2}(3x + 1){\rm{[}}{\log _3}x - 2] = 0\)

\(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}{\log _2}(3x + 1) = 0\\{\log _3}x - 2 = 0\end{array} \right.\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}3x + 1 = 1\\{\log _3}x = 2\end{array} \right.\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}x = 0(l)\\x = 9\end{array} \right. \Leftrightarrow x = 9\).

c) ĐK: x > 0

\(\begin{array}{l}
{2^{{{\log }_3}{x^2}}}{.5^{{{\log }_3}x}} = 400\\
 \Leftrightarrow {2^{2{{\log }_3}x}}{.5^{{{\log }_3}x}} = 400\\
 \Leftrightarrow {20^{{{\log }_3}x}} = {20^2}\\
 \Leftrightarrow {\log _3}x = 2\\
 \Leftrightarrow x = 9\,\,\left( n \right)
\end{array}\)

d) ĐK: x > 0

\(\begin{array}{l}
{\ln ^3}x - 3{\ln ^2}x - 4\ln x + 12 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\ln x =  - 2\\
\ln x = 3\\
\ln x = 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = {e^{ - 2}}\\
x = {e^3}\\
x = {e^2}
\end{array} \right.
\end{array}\)

-- Mod Toán 12 HỌC247