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 \(S=\left\{2;3\right\}\)

\(\Rightarrow x=2\)

đặt: \(\left\{{}\begin{matrix}2^x=t\\t>0\end{matrix}\right.\)

\(t^2-12t+32=0\Leftrightarrow t^2-2.6t+36=4\)

\(\left(t-6\right)^2=2^2\Rightarrow\left[{}\begin{matrix}t=8\\t=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Ta có : 4^x = (2^x)² .

=> 4^x - 12.2^x + 32 = 0 <=> (2^x)² - 12.2^x + 36 - 4 = 0

<=> (2^x - 6 )² - 4 = 0

<=> (2^x - 6 - 2 ).( 2^x - 6 + 2 ) = 0

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

=> \(\left[{}\begin{matrix}2^x=8\\2^x=4\end{matrix}\right.\) => \(\left[{}\begin{matrix}2^x=2^3\\2^x=2^2\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

ban coi ki laoi de coi chung de sai do cho 4x

mình giải đc rồi cảm ơn bạn nha