# Bài 69 trang 124 SGK giải tích 12 nâng cao

##### Hướng dẫn giải

a) Điều kiện: $$x> 0$$

\eqalign{ & \,{\log ^2}{x^3} - 20\log \sqrt x + 1 = 0 \Leftrightarrow {\left( {3\log x} \right)^2} - 10\log x + 1 = 0 \cr & \Leftrightarrow 9{\log ^2}x - 10\log x + 1 = 0 \Leftrightarrow \left[ \matrix{ \log x = 1 \hfill \cr \log x = {1 \over 9} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 10 \hfill \cr x = {10^{{1 \over {9}}}} = \root 9 \of {10} \hfill \cr} \right. \cr}

Vậy $$S = \left\{ {10;\root 9 \of {10} } \right\}$$
b) $$\,{{{{\log }_2}x} \over {{{\log }_4}2x}} = {{{{\log }_8}4x} \over {{{\log }_{16}}8x}}\,\,\,\,\,\left( 1 \right)$$
Điều kiện: $$x > 0$$, $$x \ne {1 \over 2},\,x \ne {1 \over 8}$$
Ta có: $${\log _4}2x = {{{{\log }_2}2x} \over {{{\log }_2}4}} = {{1 + {{\log }_2}x} \over 2}$$

\eqalign{ & {\log _8}4x = {{{{\log }_2}4x} \over {{{\log }_2}8}} = {{2 + {{\log }_2}x} \over 3} \cr & {\log _{16}}8x = {{{{\log }_2}8x} \over {{{\log }_2}16}} = {{3 + {{\log }_2}x} \over 4} \cr}

Đặt $$t = {\log _2}x$$ thì (1) thành: $${{2t} \over {1 + t}} = {{4\left( {2 + t} \right)} \over {3\left( {3 + t} \right)}} \Leftrightarrow 6t\left( {3 + t} \right) = 4\left( {1 + t} \right)\left( {2 + t} \right)$$

\eqalign{ & \Leftrightarrow 18t + 6{t^2} = 8 + 12t + 4{t^2} \Leftrightarrow 2{t^2} + 6t - 8 = 0 \Leftrightarrow \left[ \matrix{ t = 1 \hfill \cr t = - 4 \hfill \cr} \right. \cr & \left[ \matrix{ {\log _2}x = 1 \hfill \cr {\log _2}x = - 4 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 2 \hfill \cr x = {2^{ - 4}} = {1 \over {16}} \hfill \cr} \right. \cr}

Vậy $$S = \left\{ {2;{1 \over {16}}} \right\}$$
c) Điều kiện: $$x > 0$$; $$x \ne {1 \over 9},\,x \ne {1 \over 3}$$
Ta có: $${\log _{9x}}27 - {\log _{3x}}3 + {\log _9}243 = 0 \Leftrightarrow {1 \over {{{\log }_{27}}9x}} - {1 \over {{{\log }_3}3x}} + {\log _{{3^2}}}{3^5} = 0$$

\eqalign{ & \Leftrightarrow {1 \over {{{\log }_{{3^3}}}9x}} - {1 \over {1 + {{\log }_3}x}} + {5 \over 2} = 0 \cr & \Leftrightarrow {3 \over {{{\log }_3}9x}} - {1 \over {1 + {{\log }_3}x}} + {5 \over 2} = 0 \cr & \Leftrightarrow {3 \over {2 + {{\log }_3}x}} - {1 \over {1 + {{\log }_3}x}} + {5 \over 2} = 0 \cr}

Đặt $${\log _3}x = t$$
Ta có phương trình: $${3 \over {t + 2}} - {1 \over {t + 1}} + {5 \over 2} = 0$$

\eqalign{ & \Leftrightarrow 6\left( {t + 1} \right) - 2\left( {t + 2} \right) + 5\left( {t + 2} \right)\left( {t + 1} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ t = - 0,8 \hfill \cr t = - 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {\log _3}x = - 0,8 \hfill \cr {\log _3}x = - 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {3^{ - 0,8}} \hfill \cr x = {3^{ - 3}} \hfill \cr} \right. \cr}

Vậy $$S = \left\{ {{3^{ - 3}};{3^{ - 0,8}}} \right\}$$