Question

Answer 1

\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=1\\\dfrac{3}{4}x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{3}{2};-\dfrac{2}{3}\right\}\) là giá trị cần tìm => Chọn C

Answer 2

<=>\(\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

=>C

Answer 3