\(\Rightarrow x=2\)

đặt 2^xlà a thì ra rồi

x=2 hoặcx= 3

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

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

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

Vậy \(S=\left\{2;3\right\}\)

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

⇒ \(2^x.2^x-4.2^x-8.2^x+4.8=0\)

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

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 .

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

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

=>x=3 hoặc x=2

a)

\(x^2-4x+4=25\)

\(\Leftrightarrow x^2-4x-21=0\)

\(\Leftrightarrow x^2+3x-7x-21=0\)

\(\Leftrightarrow x\left(x+3\right)-7\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b)

\(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)

\(\Leftrightarrow\dfrac{x-17}{1990}-1+\dfrac{x-21}{1986}-1+\dfrac{x+1}{1004}-2=4-1-1-2\)

\(\Leftrightarrow\dfrac{x-17-1990}{1990}+\dfrac{x-21-1986}{1986}+\dfrac{x+1-2008}{1004}=0\)

\(\Leftrightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)

\(\Leftrightarrow x-2007=0\) ( Vì: \(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\ne0\))

\(\Leftrightarrow x=2007\)

c.

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

\(\Leftrightarrow\left(2^x\right)^2-12.2^x+36-4=0\)

\(\Leftrightarrow2^x-2.2^x.6+6^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6\right)^2-2^2=0\)

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

\(\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x-8=0\\2^x-4=0\end{matrix}\right.\)

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

x=3