Bài tập 2.46 trang 124 SBT Toán 12

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) \(5{x^2} - 5x - 6 = 1 \Leftrightarrow {x^2} - 5x - 6 = 0 \Leftrightarrow \left[ \begin{array}{l}

x =  - 1\\

x = 6

\end{array} \right.\)

c) 

\(\begin{array}{l}

{\left( {\frac{1}{7}} \right)^{{x^2} - 2x - 3}} = {7^{x + 1}}\\

 \Leftrightarrow {7^{ - {x^2} + 2x + 3}} = {7^{x + 1}}\\

 \Leftrightarrow  - {x^2} + 2x + 3 = x + 1\\

 \Leftrightarrow  - {x^2} + x + 2 = 0\\

 \Leftrightarrow \left[ \begin{array}{l}

x =  - 1\\

x = 2

\end{array} \right.

\end{array}\)

d) ĐKXĐ: \(x \ne 7;x \ne 3\)

\(\begin{array}{l}

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

 \Leftrightarrow {2^{\frac{{5(x + 5)}}{{x - 7}}}} = \frac{1}{4}{.5^{\frac{{3(x + 17)}}{{x - 3}}}}\\

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

 \Leftrightarrow \frac{{5x + 25 + 2x - 14}}{{x - 7}} = \frac{{3x + 51}}{{x - 3}}.{\log _2}5\\

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

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

 \Leftrightarrow (7 - 3{\log _2}5).{x^2} - 2(5 + 15{\log _2}5)x - (33 - 357{\log _2}5) = 0

\end{array}\)

Ta có:

\({\rm{\Delta '}} = 1296\log _2^25 - 2448{\log _2}5 + 256 > 0\)

Phương trình có hai nghiệm

\(x = \frac{{5 + 15{{\log }_2}5 \pm \sqrt {{\rm{\Delta '}}} }}{{7 - 3{{\log }_2}5}}\) (TMĐK)

-- Mod Toán 12