Bài tập 93 trang 131 SGK Toán 12 NC

Bài tập 93 trang 131 SGK Toán 12 NC

a) Ta có: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{32^{\frac{{x + 5}}{{x - 7}}}} = 0,{25.128^{\frac{{x + 17}}{{x - 3}}}}\\
 \Leftrightarrow {2^{\frac{{5(x + 5)}}{{x - 7}}}} = \frac{1}{4}{.2^{\frac{{7(x + 17)}}{{x - 3}}}}
\end{array}\\
\begin{array}{l}
 \Leftrightarrow {2^{\frac{{5\left( {x + 5} \right)}}{{x - 7}}}} = {2^{\frac{{7\left( {x + 17} \right)}}{{x - 3}} - 2}}\\
 \Leftrightarrow \frac{{5\left( {x + 5} \right)}}{{x - 7}} = \frac{{7\left( {x + 17} \right)}}{{x - 3}} - 2\left( 1 \right)
\end{array}
\end{array}\)

Điều kiện \(x \ne 3;x \ne 7.\)

\(\begin{array}{l}
 \Leftrightarrow 5\left( {x + 5} \right)\left( {x - 3} \right)\\
 = 7\left( {x + 17} \right)\left( {x - 7} \right) - 2\left( {x - 7} \right)\left( {x - 3} \right)
\end{array}\)

\( \Leftrightarrow 80x = 800 \Leftrightarrow x = 10\) (nhận)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{5^{x - 1}} = {10^x}{.2^{ - x}}{.5^{x + 1}}\\
 \Leftrightarrow \frac{1}{5}{.5^x} = \frac{{{{10}^x}}}{{{2^x}}}{.5.5^x}\\
 \Leftrightarrow \frac{1}{5} = {5^x}.5
\end{array}\\
{ \Leftrightarrow {5^x} = \frac{1}{{25}} \Leftrightarrow x =  - 2}
\end{array}\)

Vậy S = {-2}

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{4^x} - {3^{x - 0,5}} = {3^{x + 0,5}} - {2^{2x - 1}}\\
 \Leftrightarrow {4^x} + \frac{1}{2}{.4^x} = {3^{x - 0,5}} + {3^{x + 0,5}}
\end{array}\\
\begin{array}{l}
 \Leftrightarrow \frac{3}{2}{.4^x} = {3^{x - 0,5}}\left( {1 + 3} \right)\\
 \Leftrightarrow \frac{1}{2}{4^{x - 1}} = {3^{x - 1,5}}
\end{array}\\
\begin{array}{l}
 \Leftrightarrow {4^{x - 1,5}} = {3^{x - 0,5}}\\
 \Leftrightarrow {\left( {\frac{4}{3}} \right)^{x - 1,5}} = 1 \Leftrightarrow x - 1,5 = 0
\end{array}\\
{ \Leftrightarrow x = 1,5}
\end{array}\)

Vậy S = [1,5}

d) Đặt \(t = {3^{2x + 4}} \left( {t > 0} \right)\)

Ta có phương trình: 

\({t^2} - 12t + 28 = 1 \Leftrightarrow {t^2} - 12t + 27 = 0\)

\(\begin{array}{l}
 \Leftrightarrow \left[ \begin{array}{l}
t = 9\\
t = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{3^{2x + 4}} = 9\\
{3^{2x + 4}} = 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 4 = 2\\
2x + 2 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x =  - \frac{3}{2}
\end{array} \right.
\end{array}\)

Vậy \(S = \left\{ { - \frac{3}{2}; - 1} \right\}\)

-- Mod Toán 12