Bài 75 trang 127 SGK giải tích 12 nâng cao

LG a

$\eqalign{ {\log _3}\left( {{3^x} - 1} \right).{\log _3}\left( {{3^{x + 1}} - 3} \right) = 12; \cr}$   

Lời giải chi tiết:

Điều kiện:

$\begin{array}{l} \left\{ \begin{array}{l} {3^x} - 1 > 0\\ {3^{x + 1}} - 3 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {3^x} - 1 > 0\\ {3.3^x} - 3 > 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {3^x} - 1 > 0\\ 3\left( {{3^x} - 1} \right) > 0 \end{array} \right. \Leftrightarrow {3^x} - 1 > 0\\ \Leftrightarrow {3^x} > 1 \Leftrightarrow x > 0 \end{array}$

Ta có: $lo{g_3}\left( {{3^x} - 1} \right).lo{g_3}\left( {{3^{x + 1}} - 3} \right) = 12$ 

$\eqalign{ & \Leftrightarrow lo{g_3}\left( {{3^x} - 1} \right).lo{g_3}[3\left( {{3^x} - 1} \right)] = 12 \cr  & \Leftrightarrow lo{g_3}\left( {{3^x} - 1} \right)\left[ {1 + lo{g_3}\left( {{3^x} - 1} \right)} \right] = 12 \cr}$

$\Leftrightarrow \log _3^2\left( {{3^x} - 1} \right) + lo{g_3}\left( {{3^x} - 1} \right) - 12 = 0$

$\eqalign{ & \Leftrightarrow \left[ \matrix{ lo{g_3}\left( {{3^x} - 1} \right) = - 4 \hfill \cr  lo{g_3}\left( {{3^x} - 1} \right) = 3 \hfill \cr} \right. \cr&\Leftrightarrow \left[ \matrix{ {3^x} - 1 = 3^{-4}={1 \over {81}} \hfill \cr  {3^x} - 1 = {3^3} = 27 \hfill \cr} \right. \cr  & \Leftrightarrow \left[ \matrix{ {3^x} = {{82} \over {81}} \hfill \cr  {3^x} = 28 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {\log _3}{{82} \over {81}} \hfill \cr  x = {\log _3}28 \hfill \cr} \right. \cr}$

Vậy $S = \left\{ {{{\log }_3}28;{\log _3}{{82} \over {81}} } \right\}$

Quảng cáo

LG b

$\eqalign{ {\log _{x - 1}}4 = 1 + {\log _2}\left( {x - 1} \right); \cr}$    

Lời giải chi tiết:

Điều kiện: $0 < x - 1 \ne 1 \Leftrightarrow 1 < x \ne 2$

Ta có: ${\log _{x - 1}}4 = {1 \over {{{\log }_4}\left( {x - 1} \right)}}$

$= \frac{1}{{{{\log }_{{2^2}}}\left( {x - 1} \right)}} = \frac{1}{{\frac{1}{2}{{\log }_2}\left( {x - 1} \right)}}$

$= {2 \over {{{\log }_2}\left( {x - 1} \right)}}$.

Đặt $t = {\log _2}\left( {x - 1} \right)$

Ta có phương trình:

$\eqalign{ & {2 \over t} = 1 + t \Leftrightarrow {t^2} + t - 2 = 0 \cr  & \Leftrightarrow \left[ \matrix{ t = 1 \hfill \cr  t = - 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {\log _2}\left( {x - 1} \right) = 1 \hfill \cr  {\log _2}\left( {x - 1} \right) = - 2 \hfill \cr} \right.\cr& \Leftrightarrow \left[ \begin{array}{l}x - 1 = 2\\x - 1 = {2^{ - 2}} = \frac{1}{4}\end{array} \right. \cr&\Leftrightarrow \left[ \matrix{x = 3 \hfill \cr x = {5 \over 4} \hfill \cr} \right. (TM)\cr}$

Vậy $S = \left\{ {3;{5 \over 4}} \right\}$

LG c

$\eqalign{ 5\sqrt {{{\log }_2}\left( { - x} \right)} = {\log _2}\sqrt {{x^2}} ; \cr}$   

Lời giải chi tiết:

Điều kiện:

$\left\{ \begin{array}{l} - x > 0\\ {\log _2}\left( { - x} \right) \ge 0\\ \sqrt {{x^2}} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x < 0\\ - x \ge {2^0} = 1\\ x \ne 0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x < 0\\ x \le - 1 \end{array} \right. \Leftrightarrow x \le - 1$

$5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\sqrt {{x^2}}$

$\Leftrightarrow 5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\left| x \right|$

$\Leftrightarrow 5\sqrt {{{\log }_2}\left( { - x} \right)}  = {\log _2}\left( { - x} \right)$ (vì $x \le  - 1 \Rightarrow \left| x \right| =  - x$)

Đặt $t = {\log _2}\left( { - x} \right) \ge 0$ ta được:

$\eqalign{ & 5\sqrt t  = t \Leftrightarrow 25t = {t^2} \cr &\Leftrightarrow \left[ \matrix{ t = 0 \hfill \cr  t = 25 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {\log _2}\left( { - x} \right) = 0 \hfill \cr  lo{g_2}\left( { - x} \right) = 25 \hfill \cr} \right. \cr  & \Leftrightarrow \left[ \matrix{ x = - 1 \hfill \cr  x = - {2^{25}} \hfill \cr} \right. \cr}$

Vậy $S = \left\{ { - 1; - {2^{25}}} \right\}$

LG d

$\eqalign{ {3^{{{\log }_4} x+ {1 \over 2}}} + \,{3^{{{\log }_4} x- {1 \over 2}}} = \sqrt x . \cr}$    

Lời giải chi tiết:

Điều kiện: $x > 0$

Ta có: $\sqrt x  = \sqrt {{4^{{{\log }_4}x}}}  = {2^{{{\log }_4}x}}$

Do đó ${3^{{1 \over 2} + {{\log }_4}x}} + {3^{{{\log }_4}x - {1 \over 2}}} = \sqrt x$

$\Leftrightarrow {3^{\frac{1}{2}}}{.3^{{{\log }_4}x}} + {3^{{{\log }_4}x}}{.3^{ - \frac{1}{2}}} = {2^{{{\log }_4}x}}$

$\Leftrightarrow \left( {\sqrt 3  + {1 \over {\sqrt 3 }}} \right){3^{{{\log }_4}x}} = {2^{{{\log }_4}x}}$ 

$\eqalign{ & \Leftrightarrow {4 \over {\sqrt 3 }} = {\left( {{2 \over 3}} \right)^{{{\log }_4}x}} \cr&\Leftrightarrow {\log _4}x = {\log _{{2 \over 3}}}{4 \over {\sqrt 3 }} \cr  & \Leftrightarrow x = {4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}} \cr}$

Vậy $S = \left\{ {{4^{{{\log }_{{2 \over 3}}}{4 \over {\sqrt 3 }}}}} \right\}$

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