`@` `\text {Ans}`
`\downarrow`
\(\left(8x-16\right)\cdot\left(x-4^3\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}8x-16=0\\x-4^3=0\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}8x=16\\x=64+0\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=16\div8\\x=64\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=2\\x=64\end{matrix}\right.\)
Vậy, \(x\in\left\{2;64\right\}\)
(8\(x\) -16)(\(x-4^3\)) = 0
\(\left[{}\begin{matrix}8x-16=0\\x-4^3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}8x=16\\x=64\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=64\end{matrix}\right.\)
Vậy \(x\in\) { 2; 64}
Kurouba ryousuke
43 mà em ơi
43 = 64 mà em
Dạ e sửa rồi ạ
