Bài 93 trang 131 SGK giải tích 12 nâng cao

LG a

$\eqalign{ {32^{{{x + 5} \over {x - 7}}}} = 0,{25.128^{{{x + 17} \over {x - 3}}}}\,; \cr}$

Lời giải chi tiết:

Ta có: ${32^{{{x + 5} \over {x - 7}}}} = 0,{25.128^{{{x + 17} \over {x - 3}}}}$

$\Leftrightarrow {\left( {{2^5}} \right)^{\frac{{x + 5}}{{x - 7}}}} = \frac{1}{4}.{\left( {{2^7}} \right)^{\frac{{x + 17}}{{x - 3}}}}$

$\Leftrightarrow {2^{{{5\left( {x + 5} \right)} \over {x - 7}}}} = {2^{-2}}{.2^{{{7\left( {x + 17} \right)} \over {x - 3}}}}$

$\Leftrightarrow {2^{{{5\left( {x + 5} \right)} \over {x - 7}}}} = {2^{{{7\left( {x + 17} \right)} \over {x - 3}}-2}}$

$\Leftrightarrow {{5\left( {x + 5} \right)} \over {x - 7}} = {{7\left( {x + 17} \right)} \over {x - 3}} - 2\,\,\left( 1 \right)$

Điều kiện: $x \ne 3;\,x \ne 7.$

$(1)\Rightarrow 5\left( {x + 5} \right)\left( {x - 3} \right)$ $= 7\left( {x + 17} \right)\left( {x - 7} \right)$ $- 2\left( {x - 7} \right)\left( {x - 3} \right)$

$\Leftrightarrow 5\left( {{x^2} + 2x - 15} \right)$ $= 7\left( {{x^2} + 10x - 119} \right)$ $- 2\left( {{x^2} - 10x + 21} \right)$

$\Leftrightarrow 5{x^2} + 10x - 75$ $= 7{x^2} + 70x - 833 - 2{x^2} + 20x - 42$

$\Leftrightarrow 80x = 800$

$\Leftrightarrow x = 10$ (nhận)

Vậy $S = \left\{ {10} \right\}$

LG b

$\eqalign{ {5^{x - 1}} = {10^x}{.2^{ - x}}{.5^{x + 1}}\,; \cr}$

Lời giải chi tiết:

${5^{x - 1}} = {10^x}{.2^{ - x}}{.5^{x + 1}}$

$\Leftrightarrow {1 \over 5}{.5^x} = {{{{10}^x}} \over {{2^x}}}{.5.5^x}$

$\Leftrightarrow {1 \over 5} = {5^x}.5$

$\Leftrightarrow {5^x} = {1 \over {25}}$

$\Leftrightarrow x =  - 2$

Vậy $S = \left\{ { - 2} \right\}$

LG c

$\eqalign{ {4^x} - {3^{x - 0,5}} = {3^{x + 0,5}} - {2^{2x - 1}}\,; \cr}$

Lời giải chi tiết:

$\eqalign{ &{4^x} - {3^{x - 0,5}} = {3^{x + 0,5}} - {2^{2x - 1}}\cr& \Leftrightarrow {4^x} + {2^{2x - 1}} = {3^{x - 0,5}} + {3^{x + 0,5}}\cr& \Leftrightarrow {4^x} + {1 \over 2}{.4^x} = {3^{x - 0,5}} + 3.{3^{x - 0,5}} \cr  & \Leftrightarrow {3 \over 2}{.4^x} = {3^{x - 0,5}}\left( {1 + 3} \right) \cr}$

$\begin{array}{l} \Leftrightarrow \frac{3}{2}{.4^x} = {3^{x - 0,5}}.4\\ \Leftrightarrow {3.4^x} = {8.3^{x - 0,5}}\\ \Leftrightarrow {3.4^x} = 8.\frac{{{3^x}}}{{\sqrt 3 }}\\ \Leftrightarrow \frac{{{4^x}}}{{{3^x}}} = \frac{8}{{3\sqrt 3 }}\\ \Leftrightarrow {\left( {\frac{4}{3}} \right)^x} = {\left( {\frac{2}{{\sqrt 3 }}} \right)^3}\\ \Leftrightarrow {\left( {\frac{2}{{\sqrt 3 }}} \right)^{2x}} = {\left( {\frac{2}{{\sqrt 3 }}} \right)^3}\\ \Leftrightarrow 2x = 3 \Leftrightarrow x = \frac{3}{2} \end{array}$

Vậy $S = \left\{ {\frac{3}{2} } \right\}$

LG d

$\eqalign{ {3^{4x + 8}} - {4.3^{2x + 5}} + 28 = 2{\log _2}\sqrt 2 . \cr}$

Lời giải chi tiết:

$\begin{array}{l} \Leftrightarrow {3^{2\left( {2x + 4} \right)}} - {4.3.3^{2x + 4}} + 28 = {\log _2}{\left( {\sqrt 2 } \right)^2}\\ \Leftrightarrow {\left( {{3^{2x + 4}}} \right)^2} - {12.3^{2x + 4}} + 28 = 1\\ \Leftrightarrow {\left( {{3^{2x + 4}}} \right)^2} - {12.3^{2x + 4}} + 27 = 0 \end{array}$

Đặt $t = {3^{2x + 4}}\,\left( {t > 0} \right)$

Ta có phương trình: ${t^2} - 12t + 27 = 0$

$\eqalign{ & \Leftrightarrow \left[ \matrix{ t = 9 \hfill \cr  t = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ {3^{2x + 4}} = 9 \hfill \cr  {3^{2x + 4}} = 3 \hfill \cr} \right. \cr  & \Leftrightarrow \left[ \matrix{ 2x + 4 = 2 \hfill \cr  2x + 2 = 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = - 1 \hfill \cr  x = - {3 \over 2} \hfill \cr} \right. \cr}$

Vậy $S = \left\{ { - {3 \over 2}; - 1} \right\}$

  Loigiaihay.com