a) \(\displaystyle {\left( {\frac{3}{4}} \right)^{2x - 3}} = {\left( {\frac{4}{3}} \right)^{5 - x}} \) \(\Leftrightarrow {\left( {\frac{3}{4}} \right)^{2x - 3}} = {\left( {\frac{3}{4}} \right)^{x - 5}}\)\(\displaystyle \Leftrightarrow 2x - 3 = x - 5 \Leftrightarrow x = - 2\)
b) \(\displaystyle {5^{{x^2} - 5x - 6}} = {5^0} \Leftrightarrow {x^2} - 5x - 6 = 0\)\(\displaystyle \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - 1}\\{x = 6}\end{array}} \right.\)
c) \(\displaystyle {\left( {\frac{1}{7}} \right)^{{x^2} - 2x - 3}} = {\left( {\frac{1}{7}} \right)^{ - x - 1}}\)\(\displaystyle \Leftrightarrow {x^2} - 2x - 3 = - x - 1\) \(\displaystyle \Leftrightarrow {x^2} - x - 2 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - 1}\\{x = 2}\end{array}} \right.\)
d) \(\displaystyle {2^{5.\frac{{x + 5}}{{x - 7}}}} = {2^{ - 2}}{.5^{3.\frac{{x + 17}}{{x - 3}}}}\)\(\displaystyle \Leftrightarrow {2^{\frac{{5x + 25}}{{x - 7}} + 2}} = {5^{\frac{{3x + 51}}{{x - 3}}}}\) \(\displaystyle \Leftrightarrow {2^{\frac{{7x + 11}}{{x - 7}}}} = {5^{\frac{{3x + 51}}{{x - 3}}}}\)
Lấy logarit cơ số 2 cả hai vế, ta được:
\(\displaystyle \frac{{7x + 11}}{{x - 7}} = \frac{{3x + 51}}{{x - 3}}{\log _2}5\)\(\displaystyle \Leftrightarrow 7{x^2} - 10x - 33\)\(\displaystyle = (3{x^2} + 30x - 357){\log _2}5\) (với \(\displaystyle x \ne 7,x \ne 3\))
\(\displaystyle \Leftrightarrow (7 - 3{\log _2}5){x^2} - 2(5 + 15{\log _2}5)x\)\(\displaystyle - (33 - 357{\log _2}5) = 0\)
Ta có: \(\displaystyle \Delta ' = {(5 + 15{\log _2}5)^2}\)\(\displaystyle + (7 - 3{\log _2}5)(33 - 357{\log _2}5)\)\(\displaystyle = 1296\log _2^25 - 2448{\log _2}5 + 256 > 0\)
Phương trình đã cho có hai nghiệm: \(\displaystyle x = \frac{{5 + 15{{\log }_2}5 \pm \sqrt {\Delta '} }}{{7 - 3{{\log }_2}5}}\), đều thỏa mãn điều kiện \(\displaystyle x \ne 7,x \ne 3\)
