Hai phương trình \(2{\log _5}\left( {3x - 1} \righ...

C

- Hướng dẫn giải

Phương pháp giải:

Giải từng phương trình, sử dụng các công thức

\(\begin{array}{l}{\log _{{a^n}}}{f^m}\left( x \right) = \frac{m}{n}{\log _a}f\left( x \right)\\{\log _a}f\left( x \right) + {\log _a}g\left( x \right) = {\log _a}\left[ {f\left( x \right).g\left( x \right)} \right]\\{\log _a}f\left( x \right) - {\log _a}g\left( x \right) = {\log _a}\frac{{f\left( x \right)}}{{g\left( x \right)}}\\\left( {0 < a \ne 1;\,\,f\left( x \right) > 0;\,\,g\left( x \right) > 0} \right)\end{array}\)

Giải chi tiết:

Giải phương trình: \(2{\log _5}\left( {3x - 1} \right) + 1 = {\log _{\sqrt[3]{5}}}\left( {2x + 1} \right)\)

ĐK: \(\left\{ \begin{array}{l}3x - 1 > 0\\2x + 1 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > \frac{1}{3}\\x >  - \frac{1}{2}\end{array} \right. \Leftrightarrow x > \frac{1}{3}\)

\(\begin{array}{l}\,\,\,\,\,\,2{\log _5}\left( {3x - 1} \right) + 1 = {\log _{\sqrt[3]{5}}}\left( {2x + 1} \right)\\ \Leftrightarrow 2.{\log _5}\left( {3x - 1} \right) + 1 = 3.{\log _5}\left( {2x + 1} \right)\\ \Leftrightarrow {\log _5}{\left( {3x - 1} \right)^2} - {\log _5}{\left( {2x + 1} \right)^3} =  - 1\\ \Leftrightarrow {\log _5}\frac{{{{\left( {3x - 1} \right)}^2}}}{{{{\left( {2x + 1} \right)}^3}}} =  - 1\\ \Leftrightarrow \frac{{{{\left( {3x - 1} \right)}^2}}}{{{{\left( {2x + 1} \right)}^3}}} = \frac{1}{5}\\ \Leftrightarrow 5{\left( {3x - 1} \right)^2} = {\left( {2x + 1} \right)^3}\\ \Leftrightarrow 5\left( {9{x^2} - 6x + 1} \right) = 8{x^3} + 12{x^2} + 6x + 1\\ \Leftrightarrow 8{x^3} - 33{x^2} + 36x - 4 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{1}{8}\,\,\left( {ktm} \right)\\x = 2\,\,\,\left( {tm} \right)\end{array} \right. \Rightarrow {x_1} = 2\end{array}\)

Giải phương trình: \({\log _2}\left( {{x^2} - 2x - 8} \right) = 1 - {\log _{\frac{1}{2}}}\left( {x + 2} \right)\)

ĐK : \(\left\{ \begin{array}{l}{x^2} - 2x - 8 > 0\\x + 2 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x > 4\\x <  - 2\end{array} \right.\\x >  - 2\end{array} \right. \Leftrightarrow x > 4\)

\(\begin{array}{l}{\log _2}\left( {{x^2} - 2x - 8} \right) = 1 - {\log _{\frac{1}{2}}}\left( {x + 2} \right)\\ \Leftrightarrow {\log _2}\left( {{x^2} - 2x - 8} \right) = 1 + {\log _2}\left( {x + 2} \right)\\ \Leftrightarrow {\log _2}\left( {{x^2} - 2x - 8} \right) - {\log _2}\left( {x + 2} \right) = 1\\ \Rightarrow {\log _2}\frac{{{x^2} - 2x - 8}}{{x + 2}} = 1\\ \Leftrightarrow \frac{{{x^2} - 2x - 8}}{{x + 2}} = 2\\ \Leftrightarrow {x^2} - 2x - 8 = 2x + 4\\ \Leftrightarrow {x^2} - 4x - 12 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 6\,\,\,\,\,\,\,\left( {tm} \right)\\x =  - 2\,\,\left( {ktm} \right)\end{array} \right. \Rightarrow {x_2} = 6\end{array}\)

Vậy \({x_1} + {x_2} = 2 + 6 = 8\)

Chọn C.