Bài tập 2.46 trang 124 SBT Toán 12

a)

\(\begin{array}{l}
{\left( {0,75} \right)^{2x - 3}} = {\left( {1\frac{1}{3}} \right)^{5 - x}}\\
 \Leftrightarrow {\left( {\frac{3}{4}} \right)^{2x - 3}} = {\left( {\frac{4}{3}} \right)^{5 - x}} = {\left( {\frac{3}{4}} \right)^{x - 5}}\\
 \Leftrightarrow 2x - 3 = x - 5\\
 \Leftrightarrow x =  - 2
\end{array}\)

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)

\(\displaystyle {32^{\frac{{x + 5}}{{x - 7}}}} = 0,{25.125^{\frac{{x + 17}}{{x - 3}}}}\)

\(\begin{array}{l}
\Leftrightarrow {\left( {{2^5}} \right)^{\frac{{x + 5}}{{x - 7}}}} = \frac{1}{4}.{\left( {{5^3}} \right)^{\frac{{x + 17}}{{x - 3}}}}\\
\Leftrightarrow {4.2^{5.\frac{{x + 5}}{{x - 7}}}} = {5^{3.\frac{{x + 17}}{{x - 3}}}}\\
\Leftrightarrow {2^2}{.2^{\frac{{5x + 25}}{{x - 7}}}} = {5^{\frac{{3x + 51}}{{x - 3}}}}\\
\Leftrightarrow {2^{2 + \frac{{5x + 25}}{{x - 7}}}} = {5^{\frac{{3x + 51}}{{x - 3}}}}
\end{array}\)

\(\displaystyle  \Leftrightarrow {2^{\frac{{7x + 11}}{{x - 7}}}} = {5^{\frac{{3x + 51}}{{x - 3}}}}\)

Lấy logarit cơ số 2 cả hai vế, ta được:

\({\log _2}\left( {{2^{\frac{{7x + 11}}{{x - 7}}}}} \right) = {\log _2}\left( {{5^{\frac{{3x + 51}}{{x - 3}}}}} \right)\)

\(\displaystyle \Leftrightarrow \frac{{7x + 11}}{{x - 7}} = \frac{{3x + 51}}{{x - 3}}{\log _2}5\)

\(\Rightarrow \left( {7x + 11} \right)\left( {x - 3} \right) \) \(= \left( {3x + 51} \right)\left( {x - 7} \right){\log _2}5\)

\(\displaystyle  \Leftrightarrow 7{x^2} - 10x - 33\)\(\displaystyle  = (3{x^2} + 30x - 357){\log _2}5\)  (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 \(\displaystyle x \ne 7,x \ne 3\)

-- Mod Toán 12 HỌC247