a: \(2\left|3-2x\right|+\dfrac{1}{2}=\dfrac{5}{2}\)
=>\(2\left|2x-3\right|=2\)
=>|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
b: \(x^2\left(2^x-6\right)-2x^3=0\)
=>\(x^2\left(2^x-6-2x\right)=0\)
=>\(\left[{}\begin{matrix}x^2=0\\2^x-6-2x=0\end{matrix}\right.\Leftrightarrow x=0\)
Đúng 1
Bình luận (0)