- LG a
- LG b
- LG c
- LG d
Giải phương trình:
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 \[S = \left\{ { - {3 \over 2}; - 1} \right\}\]