Bài tập 69 trang 124 SGK Toán 12 NC

a) Điều kiện x > 0

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\log ^2}{x^3} - 20\log \sqrt x  + 1 = 0\\
 \Leftrightarrow {(3\log x)^2} - 10\log x + 1 = 0
\end{array}\\
\begin{array}{l}
 \Leftrightarrow 9{\log ^2}x - 10\log x + 1 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\log x = 1}\\
{\log x = \frac{1}{9}}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 10}\\
{x = {{10}^{\frac{1}{9}}} = \sqrt[9]{{10}}}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy \(S= \left\{ {10;\sqrt[9]{{10}}} \right\}\)

b)

\(\frac{{{{\log }_2}x}}{{{{\log }_4}2x}} = \frac{{{{\log }_8}4x}}{{{{\log }_{16}}8x}}\) (1)

Điều kiện: \(x > 0,x \ne \frac{1}{2},x \ne \frac{1}{8}\)

Ta có: 

\(\begin{array}{*{20}{l}}
{{{\log }_4}2x = \frac{{{{\log }_2}2x}}{{{{\log }_2}4}} = \frac{{1 + {{\log }_2}x}}{2}}\\
{{{\log }_8}4x = \frac{{{{\log }_2}4x}}{{{{\log }_2}8}} = \frac{{2 + {{\log }_2}x}}{3}}\\
{{{\log }_{16}}8x = \frac{{{{\log }_2}8x}}{{{{\log }_2}16}} = \frac{{3 + {{\log }_2}x}}{4}}
\end{array}\)

Đặt \(t = {\log _2}x\) thì (1) thành:

\(\begin{array}{l}
\frac{{2t}}{{1 + t}} = \frac{{4(2 + t)}}{{3(3 + t)}}\\
 \Leftrightarrow 6t(3 + t) = 4(1 + t)(2 + t)\\
\begin{array}{*{20}{l}}
\begin{array}{l}
 \Leftrightarrow 18t + 6{t^2} = 8 + 12t + 4{t^2}\\
 \Leftrightarrow 2{t^2} + 6t - 8 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{t = 1}\\
{t =  - 4}
\end{array}} \right.
\end{array}\\
{\left[ {\begin{array}{*{20}{l}}
{{{\log }_2}x = 1}\\
{{{\log }_2}x =  - 4}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 2}\\
{x = {2^{ - 4}} = \frac{1}{{16}}}
\end{array}} \right.}
\end{array}
\end{array}\)

Vậy S = {2; 1/16}

c) Điều kiện: \(x > 0,x \ne \frac{1}{9},x \ne \frac{1}{3}\)

Ta có:

\(\begin{array}{l}
{\log _{9x}}27 - {\log _{3x}}3 + {\log _9}243 = 0\\
 \Leftrightarrow \frac{1}{{{{\log }_{27}}9x}} - \frac{1}{{{{\log }_3}3x}} + {\log _{{3^2}}}{3^5} = 0\\
\begin{array}{*{20}{l}}
{ \Leftrightarrow \frac{1}{{{{\log }_{{3^3}}}9x}} - \frac{1}{{1 + {{\log }_3}x}} + \frac{5}{2} = 0}\\
{ \Leftrightarrow \frac{3}{{{{\log }_3}9x}} - \frac{1}{{1 + {{\log }_3}x}} + \frac{5}{2} = 0}\\
{ \Leftrightarrow \frac{3}{{2 + {{\log }_3}x}} - \frac{1}{{1 + {{\log }_3}x}} + \frac{5}{2} = 0}
\end{array}
\end{array}\)

Đặt log₃ x = t

Ta có phương trình \(\frac{3}{{t + 2}} - \frac{1}{{t + 1}} + \frac{5}{2} = 0\)

\(\begin{array}{*{20}{l}}
{ \Leftrightarrow 6(t + 1) - 2(t + 2) + 5(t + 2)(t + 1) = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{t =  - 0,8}\\
{t =  - 3}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_3}x =  - 0,8}\\
{{{\log }_3}x =  - 3}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = {3^{ - 0,8}}}\\
{x = {3^{ - 3}}}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy \(S = \left\{ {{3^{ - 3}};{3^{ - 0,8}}} \right\}\).

-- Mod Toán 12 HỌC247