a) Đặt \(t = {81^{{{\cos }^2}x}}(1 \le t \le 81)\)
Khi đó: \({81^{{{\sin }^2}x}} = {81^{1- {{\cos }^2}x}} = {{81} \over t}\)
Phương trình trở thành:
\(\eqalign{
& {{81} \over t} + t = 30 \Leftrightarrow {t^2} - 30t + 81 = 0 \cr
& \Leftrightarrow \left[ \matrix{
t = 27 \hfill \cr
t = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{3^{4{{\cos }^2}x}} = {3^3} \hfill \cr
{3^{4{{\cos }^2}x}} = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
4{\cos ^2}x = 3 \hfill \cr
4{\cos ^2}x = 1 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
2(1 + \cos 2x) = 3 \hfill \cr
2(1 + \cos 2x) = 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
\cos 2x = {1 \over 2} \hfill \cr
\cos 2x = - {1 \over 2} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = \pm {\pi \over 6} + k\pi \hfill \cr
x = \pm {\pi \over 3} + k\pi \hfill \cr} \right. \cr} \)
b) Ta có:
\(\eqalign{
& {\log _3}(\log _{{1 \over 2}}^2x - 3{\log _{{1 \over 2}}}x + 5) = 2 \cr&\Leftrightarrow \log _{{1 \over 2}}^2x - 3{\log _{{1 \over 2}}}x + 5 = 9 \cr
& \Leftrightarrow \log _{{1 \over 2}}^2x - 3{\log _{{1 \over 2}}} - 4 = 0 \cr
& \Leftrightarrow \left[ \matrix{
{\log _{{1 \over 2}}}x = - 1 \hfill \cr
{\log _{{1 \over 2}}}x = 4 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 2 \hfill \cr
x = {1 \over {16}} \hfill \cr} \right. \cr} \)
Vậy \(S = {\rm{\{ }}{1 \over {16}};\,2\} \)
c) Điều kiện: x > 0
\(\eqalign{
& {4^{{{\log }x} + 1}} - {6^{{{\log }x}}} - {2.3^{\log {x^2} + 2}} = 0 \cr
& \Leftrightarrow {4.4^{\log x}} - {6^{\log x}} - {18.9^{\log x}} = 0 \cr} \)
Chia hai vế phương trình 4logx ta được:
\(4 - {({3 \over 2})^{\log x}} - 18.{({9 \over 4})^{\log x}} = 0\)
Đặt \(t = {({3 \over 2})^{\log x}}\,\,(t > 0)\) ta có phương trình:
\(18{t^2} + t - 4 = 0 \Leftrightarrow \left[ \matrix{ t = {4 \over 9} \hfill \cr t = - {1 \over 2}\,\,(loai) \hfill \cr} \right.\)
\(\eqalign{
& t = {4 \over 9} \Leftrightarrow {({3 \over 2})^{\log x}} = {({3 \over 2})^{-2}} \Leftrightarrow \log x = - 2 \cr
& \Leftrightarrow x = {10^{ - 2}} = {1 \over {100}} \cr} \)
d) Điều kiện: x > 0; y > 0
\(\eqalign{
& {2^x}{8^{ - y}} = 2\sqrt 2 \Leftrightarrow {2^{x - 3y}} = {2^{{3 \over 2}}} \Leftrightarrow x - 3y = {3 \over 2}\,\,\,\,\,(1) \cr
& {\log _9}{1 \over x} + {1 \over 2} = {1 \over 2}{\log _3}(9y) \cr&\Leftrightarrow {1 \over 2}{\log _3}{1 \over x} + {1 \over 2} = {1 \over 2}{\log _3}(9y) \cr
& \Leftrightarrow {\log _3}{3 \over x} = {\log _3}(9y) \Leftrightarrow {3 \over x} = 9y \Leftrightarrow xy = {1 \over 3}\,\,\,\,\,\,\,\,\,\,\,\,(2) \cr} \)
Từ (1) và (2) ta có hệ phương trình
\(\left\{ \matrix{ x - 3y = {3 \over 2} \hfill \cr xy = {1 \over 3} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = {3 \over 2} + 3y \hfill \cr ({3 \over 2} + 3y)y = {1 \over 3} \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
x = {3 \over 2} + 3y \hfill \cr
3{y^2} + {3 \over 2}y - {1 \over 3} = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x = 2 \hfill \cr
y = {1 \over 6} \hfill \cr} \right.\)
Vậy \(S = {\rm{\{ }}(2,\,{1 \over 6}){\rm{\} }}\)
