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 \(S = \left\{ { – {3 \over 2}; – 1} \right\}\)

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