Question

giải pt sau: 

\(2^{2x-3}-3.2^{x-2}+1=0\)

Answer 1

\(\Leftrightarrow2^{-3}.2^{2x}-3.2^{-2}.2^x+1=0\)

\(\Leftrightarrow\dfrac{1}{8}2^{2x}-\dfrac{3}{4}2^x+1=0\)

Đặt \(2^x=t>0\)

\(\Rightarrow\dfrac{1}{8}t^2-\dfrac{3}{4}t+1=0\Rightarrow\left[{}\begin{matrix}t=4\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)