Question

tìm x biết

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

Answer 1

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

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

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

Answer 2

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 .

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