a.
\(9^{2x-1}-3.6^{2x-1}+2.4^{2x-1}=0\)
\(\Leftrightarrow\left(\dfrac{9}{4}\right)^{2x-1}-3\left(\dfrac{6}{4}\right)^{2x-1}+2=0\)
\(\Leftrightarrow\left(\dfrac{3}{2}\right)^{2.\left(2x-1\right)}-\left(\dfrac{3}{2}\right)^{2x-1}+2=0\)
Đặt \(\left(\dfrac{3}{2}\right)^{2x-1}=t>0\)
\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(\dfrac{3}{2}\right)^{2x-1}=1\\\left(\dfrac{3}{2}\right)^{2x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=log_{\dfrac{3}{2}}2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{2}log_{\dfrac{3}{2}}2+\dfrac{1}{2}\end{matrix}\right.\)
b.
\(2^{x^2-x}-2^{2+x-x^2}=3\)
\(\Leftrightarrow2^{x^2-x}-4.2^{x-x^2}=3\)
Đặt \(2^{x^2-x}=t>0\Rightarrow2^{x-x^2}=\dfrac{1}{t}\)
Pt trở thành:
\(t-\dfrac{4}{t}=3\Rightarrow t^2-3t-4=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=4\end{matrix}\right.\)
\(\Rightarrow2^{x^2-x}=4\)
\(\Leftrightarrow x^2-x=2\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
