Bài 93 trang 131 SGK giải tích 12 nâng cao

Câu hỏi: Giải phương trình:

Câu 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 \(S = \left\{ {10} \right\}\)

Câu 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 \(S = \left\{ { - 2} \right\}\)

Câu 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 \(S = \left\{ {\frac{3}{2} } \right\}\)

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