Question

[Hoàng Ngọc Quang]

a)\(2\cdot\text{|}3-2x\text{|}+\dfrac{1}{2}=\dfrac{5}{2}\)               b)\(x^2\cdot\left(2^x-6\right)-2x^3=0\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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: \(x^2\left(2^x-6\right)-2x^3=0\)

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

=>\(\left[{}\begin{matrix}x^2=0\\2^x-6-2x=0\end{matrix}\right.\Leftrightarrow x=0\)