Ta có : \(4^x-12.2^x+32=0\)
\(\Leftrightarrow2^x.2^x-4.2^x-8.2^x+4.8=0\)
\(\Leftrightarrow2^x.\left(2^x-4\right)-8\left(2^x-4\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x-2^3=0\\2^x-2^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=2^3\\2^x=2^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy x = 3 hoặc x = 2
\(4^x-12.2^x+32=0\left(1\right)\) đăt \(t=2^x\Rightarrow t>0\)
(1) \(\Leftrightarrow\)\(\left(2^x\right)^2-12.2^x+32=0\)
\(\Leftrightarrow t^2-12t+32=0\)
\(\Leftrightarrow\) (t-8) (t-4) =0 \(\Rightarrow\left[{}\begin{matrix}t=4\\t=8\end{matrix}\right.\)
- t = 4 \(\Rightarrow2^x=2^2\Rightarrow x=2\)
- t = 8 \(\Rightarrow2^x=2^3\Rightarrow x=3\)
vậy pt có 2 nghiệm x =2 và x=3
Đặt: \(\left\{{}\begin{matrix}2^x=t\\t>0\end{matrix}\right.\)
Thay t vào biểu thức ta được:
\(t^2-12.t+32< =>t^2-2.6t+36=4\\ < =>\left(t-6\right)^2=2^2\\ =>\left[{}\begin{matrix}t=8\\t=4\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;3\right\}\)