Bài tập 2.47 trang 124 SBT Toán 12

a)

\(\displaystyle {16.2^x} + {4.2^x} = {5.5^x} + {3.5^x}\)\(\displaystyle  \Leftrightarrow {20.2^x} = {8.5^x}\) \( \Leftrightarrow \frac{{{2^x}}}{{{5^x}}} = \frac{8}{{20}} = \frac{2}{5}\) \(\displaystyle  \Leftrightarrow {\left( {\frac{2}{5}} \right)^x} = {\left( {\frac{2}{5}} \right)^1} \Leftrightarrow x = 1\)

b) 

\(\displaystyle {5^{2x}} - {7^x} - {5^{2x}}.17 + {7^x}.17 = 0\)

\(\begin{array}{l}
\Leftrightarrow \left( {{7^x}.17 - {7^x}} \right) - \left( {{5^{2x}}.17 - {5^{2x}}} \right) = 0\\
\Leftrightarrow {7^x}\left( {17 - 1} \right) - {5^{2x}}\left( {17 - 1} \right) = 0\\
\Leftrightarrow {7^x}.16 - {5^{2x}}.16 = 0\\
\Leftrightarrow {7^x} - {5^{2x}} = 0
\end{array}\)

\(\displaystyle \Leftrightarrow {7^x} = {5^{2x}}\) \(\Leftrightarrow \frac{{{7^x}}}{{{5^{2x}}}} = 1 \Leftrightarrow \frac{{{7^x}}}{{{{25}^x}}} = 1\) \(\displaystyle  \Leftrightarrow {\left( {\frac{7}{{25}}} \right)^x} = {\left( {\frac{7}{{25}}} \right)^0} \Leftrightarrow x = 0\)

c) 

Chia hai vế cho \(\displaystyle {12^x}({12^x} > 0)\), ta được:

\(\begin{array}{l}
4.\frac{{{9^x}}}{{{{12}^x}}} + 1 - 3.\frac{{{{16}^x}}}{{{{12}^x}}} = 0\\
\Leftrightarrow 4.{\left( {\frac{9}{{12}}} \right)^x} + 1 - 3.{\left( {\frac{{16}}{{12}}} \right)^x} = 0\\
\Leftrightarrow 4.{\left( {\frac{3}{4}} \right)^x} + 1 - 3.{\left( {\frac{4}{3}} \right)^x} = 0
\end{array}\)

Đặt  \(\displaystyle t = {\left( {\frac{3}{4}} \right)^x} > 0\), ta có phương trình: \(\displaystyle 4t + 1 - \frac{3}{t} = 0\)\(\displaystyle  \Leftrightarrow 4{t^2} + t - 3 = 0\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}t =  - 1\left( {KTM} \right)\\t = \frac{3}{4}\left( {TM} \right)\end{array} \right.\)

Do đó \({\left( {\frac{3}{4}} \right)^x} = \frac{3}{4} \)\(\displaystyle  \Leftrightarrow {\left( {\frac{3}{4}} \right)^x} = {\left( {\frac{3}{4}} \right)^1} \Leftrightarrow x = 1\) .

Vậy \(\displaystyle x = 1\).

d) 

\(\begin{array}{l}
- {8^x} + {2.4^x} + {2^x} - 2 = 0\\
\Leftrightarrow - {\left( {{2^3}} \right)^x} + 2.{\left( {{2^2}} \right)^x} + {2^x} - 2 = 0\\
\Leftrightarrow - {2^{3x}} + {2.2^{2x}} + {2^x} - 2 = 0\\
\Leftrightarrow - {\left( {{2^x}} \right)^3} + 2.{\left( {{2^x}} \right)^2} + {2^x} - 2 = 0
\end{array}\)

Đặt \(\displaystyle t = {2^x}(t > 0)\) , ta có phương trình:

\(\displaystyle  - {t^3} + 2{t^2} + t - 2 = 0\)\(\displaystyle  \Leftrightarrow (t - 1)(t + 1)(2 - t) = 0\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}t = 1\left( {TM} \right)\\t =  - 1\left( {KTM} \right)\\t = 2\left( {TM} \right)\end{array} \right.\)

Do đó \(\displaystyle \left[ \begin{array}{l}{2^x} = 1\\{2^x} = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 1\end{array} \right.\)

Vậy phương trình có nghiệm \(\displaystyle x = 1\), \(\displaystyle x = 0\).

-- Mod Toán 12 HỌC247