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Bài 93. Giải phương trình:
\(\eqalign{
& a)\,{32^{{{x + 5} \over {x – 7}}}} = 0,{25.128^{{{x + 17} \over {x – 3}}}}\,; \cr
& c)\,{4^x} – {3^{x – 0,5}} = {3^{x + 0,5}} – {2^{2x – 1}}\,; \cr} \)
\(\eqalign{
& b)\,{5^{x – 1}} = {10^x}{.2^{ – x}}{.5^{x + 1}}\,; \cr
& d)\,{3^{4x + 8}} – {4.3^{2x + 5}} + 28 = 2{\log _2}\sqrt 2 . \cr} \)
a) Ta có: \({32^{{{x + 5} \over {x – 7}}}} = 0,{25.128^{{{x + 17} \over {x – 3}}}} \Leftrightarrow {2^{{{5\left( {x + 5} \right)} \over {x – 7}}}} = {1 \over 4}{.2^{{{7\left( {x + 17} \right)} \over {x – 3}}}}\)
\( \Leftrightarrow {2^{{{5\left( {x + 5} \right)} \over {x – 7}}}} = {2^{{{7\left( {x + 17} \right)} \over {x – 3}}-2}} \Leftrightarrow {{5\left( {x + 5} \right)} \over {x – 7}} = {{7\left( {x + 17} \right)} \over {x – 3}} – 2\,\,\left( 1 \right)\)
Điều kiện: \(x \ne 3;\,x \ne 7.\)
(1) \( \Leftrightarrow 5\left( {x + 5} \right)\left( {x – 3} \right) = 7\left( {x + 17} \right)\left( {x – 7} \right) – 2\left( {x – 7} \right)\left( {x – 3} \right)\)
\( \Leftrightarrow 80x = 800 \Leftrightarrow x = 10\) (nhận)
Vậy \(S = \left\{ {10} \right\}\)
\(b)\,{5^{x – 1}} = {10^x}{.2^{ – x}}{.5^{x + 1}} \Leftrightarrow {1 \over 5}{.5^x} = {{{{10}^x}} \over {{2^x}}}{.5.5^x} \Leftrightarrow {1 \over 5} = {5^x}.5 \Leftrightarrow {5^x} = {1 \over {25}} \Leftrightarrow x = – 2\)
Vậy \(S = \left\{ { – 2} \right\}\)
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\(\eqalign{
& c)\,\,{4^x} – {3^{x – 0,5}} = {3^{x + 0,5}} – {2^{2x – 1}} \Leftrightarrow {4^x} + {1 \over 2}{.4^x} = {3^{x – 0,5}} + {3^{x + 0,5}} \cr
& \,\, \Leftrightarrow {3 \over 2}{.4^x} = {3^{x – 0,5}}\left( {1 + 3} \right) \Leftrightarrow {1 \over 2}{.4^{x – 1}} = {3^{x – 1,5}} \cr
& \,\, \Leftrightarrow {4^{x – 1,5}} = {3^{x – 1,5}} \Leftrightarrow {\left( {{4 \over 3}} \right)^{x – 1,5}} = 1 \Leftrightarrow x – 1,5 = 0 \cr
& \,\,\, \Leftrightarrow x = 1,5 \cr} \)
Vậy \(S = \left\{ {1,5} \right\}\)
d) Đặt \(t = {3^{2x + 4}}\,\left( {t > 0} \right)\)
Ta có phương trình: \({t^2} – 12t + 28 = 1 \Leftrightarrow {t^2} – 12t + 27 = 0\)
\(\eqalign{
& \Leftrightarrow \left[ \matrix{
t = 9 \hfill \cr
t = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{3^{2x + 4}} = 9 \hfill \cr
{3^{2x + 4}} = 3 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
2x + 4 = 2 \hfill \cr
2x + 2 = 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = – 1 \hfill \cr
x = – {3 \over 2} \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ { – {3 \over 2}; – 1} \right\}\)