Question

[Lê Mỹ Tâm]

Tìm x biết :

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

giúp mik với

Answer 1

[Trần Kiều Anh]

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

Answer 2

[nguyen ngoc song thuy]

\(4^x-12.2^x+32=0\left(1\right)\) 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