Question

Tìm x khi:

(1/2.x -3) . (2/3.x + 1/2) = 0

Answer 1

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}=6\\x=-\dfrac{1}{2}:\dfrac{2}{3}=-\dfrac{3}{4}\end{matrix}\right.\)

KL: Vậy để \(\left(\dfrac{1}{2}x-3\right)\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)=0\) thì \(x\in\left\{6;-\dfrac{3}{4}\right\}\)

Answer 2

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

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\\dfrac{2}{3}x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3\\\dfrac{2}{3}x=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-\dfrac{3}{4}\end{matrix}\right.\)