Question

\((3^x - 27).(2\sqrt{x}-4)=0\)

Answer 1

\(\Rightarrow\left[{}\begin{matrix}3^x-27=0\\2\sqrt{x}-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3^x=27\\2\sqrt{x}=4\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

\(\Rightarrow x\in\left(3;4\right)\)

Answer 2

`(3^x -27)(2sqrt{x} -4) =0`

`Đk : x in Z ; x >=0`

`=> `\(\left[{}\begin{matrix}3^x-27=0\\2\sqrt{x}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3^x=27\\2\left(\sqrt{x}-2\right)=0\end{matrix}\right.\)

`<=> [(3^x = 3^3),(sqrt{x}-2 =0):}`

`=> [(x=3),(sqrt{x}=2):}`

`=>  [(x=3(t//m) ) , (x=4 (t//m) ) :}`

Vậy `x in {3;4}`