Ta có : \(4^x+12.2^x+32=0\)

<=> \(\left(2^x\right)^2+2.6.2^x+36-4=0\)

<=> \(\left(2^x+6\right)^2-4=0\)

<=> \(\left(2^x+6+2\right)\left(2^x+6-2\right)=0\)

<=> \(\left(2^x+8\right)\left(2^x+4\right)=0\)

<=> \(\left[{}\begin{matrix}2^x+8=0\\2^x+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}2^x=-8\\2^x=-8\end{matrix}\right.\) ( Vô lý )

Vậy phương trình vô nghiệm .

đặt 2^xlà a thì ra rồi

x=2 hoặcx= 3

=>(2^x)^2-12*2^x+32=0

=>(2^x-4)(2^x-8)=0

=>x=3 hoặc x=2

x=3

\(4^x-12.2^x+32=0\)

⇒ \(2^x.2^x-4.2^x-8.2^x+4.8=0\)

⇒ \(\left[{}\begin{matrix}2^x=2^2\\2^x=2^3\end{matrix}\right.\)

\(4^x-12.2^x+32=0\Leftrightarrow\left(2^x\right)^2-2.6.2^x+6^2-4=0\Leftrightarrow\left(2^x-6\right)^2-2^2=0\)

\(\Leftrightarrow\left(2^x-6-2\right)\left(2^x-6+2\right)=0\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\Leftrightarrow\orbr{\begin{cases}2^x-8=0\\2^x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2^x=8\\2^x=4\end{cases}\Leftrightarrow\orbr{\begin{cases}2^x=2^3\\2^x=2^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

Vậy \(S=\left\{2;3\right\}\)

\(\Rightarrow x=2\)