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Bài 74.
\(\eqalign{
& a)\,{\log _2}\left( {3 – x} \right) + {\log _2}\left( {1 – x} \right) = 3; \cr
& c)\,{7^{\log x}} – {5^{\log x + 1}} = {3.5^{\log x – 1}} – 13.{7^{\log x – 1}} \cr} \)
\(\eqalign{
& b)\,{\log _2}\left( {9 – {2^x}} \right) = {10^{\log \left( {3 – x} \right)}} \cr
& d)\,{6^x} + {6^{x + 1}} = {2^x} + {2^{x + 1}} + {2^{x + 2}} \cr} \)
a) Điều kiện: \(x < 1\)
\(\eqalign{
& \,\,\,\,{\log _2}\left( {3 – x} \right) + {\log _2}\left( {1 – x} \right) = 3 \Leftrightarrow {\log _2}\left( {3 – x} \right)\left( {1 – x} \right) = 3 \cr
& \Leftrightarrow \left( {3 – x} \right)\left( {1 – x} \right) = 8 \Leftrightarrow {x^2} – 4x – 5 = 0\left[ \matrix{
x = – 1 \hfill \cr
x = 5\,\,\left( \text{loại} \right) \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ { – 1} \right\}\)
b) Điều kiện:
\(\left\{ \matrix{
3 – x > 0 \hfill \cr
9 – {2^x} > 0 \hfill \cr} \right. \Leftrightarrow x < 3\)
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\(\eqalign{
& \,\,\,\,{\log _2}\left( {9 – {2^x}} \right) = {10^{\log \left( {3 – x} \right)}} \Leftrightarrow {\log _2}\left( {9 – {2^x}} \right) = 3 – x \Leftrightarrow 9 – {2^x} = {2^{3 – x}} \cr
& \Leftrightarrow 9 – {2^x} = {8 \over {{2^x}}} \Leftrightarrow {4^x} = {9.2^x} – 8 = 0 \Leftrightarrow \left[ \matrix{
{2^x} = 1 \hfill \cr
{2^x} = 8 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 0 \hfill \cr
x = 3\,\,\left( \text{loại} \right) \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ 0 \right\}\)
c) Điều kiện: \(x > 0\)
\(\eqalign{
& \Leftrightarrow {20.7^{\lg x – 1}} = {28.5^{\lg x – 1}} \cr
& \Leftrightarrow {\left( {{7 \over 8}} \right)^{\lg x – 1}} = {7 \over 8} \cr
& \Leftrightarrow \lg x – 1 = 1 \Leftrightarrow \lg x = 2 \Leftrightarrow x = 100 \cr} \)
Vậy \(S = \left\{ {100} \right\}\)
d) Ta có:
\(\eqalign{
& {6^x} + {6^{x + 1}} = {2^x} + {2^{x + 1}} + {2^{x + 2}} \cr
& \Leftrightarrow {6^x}\left( {1 + 6} \right) = {2^x}\left( {1 + 2 + {2^2}} \right) \cr
& \Leftrightarrow {3^x} = 1 \cr
& \Leftrightarrow x = 0 \cr} \)
Vậy \(S = \left\{ 0 \right\}\)