Question

(9/4-2x)(3x+7/8)=0

Answer 1

\(\left(\dfrac{9}{4}-2x\right).\left(3x+\dfrac{7}{8}\right)=0\)

⇔\(\dfrac{9}{4}-2x=0\) hay \(3x+\dfrac{7}{8}=0\)

⇔\(2x=\dfrac{9}{4}\) hay \(3x=-\dfrac{7}{8}\)

⇔\(x=\dfrac{9}{8}\) hay \(x=-\dfrac{7}{24}\)

Answer 2

\(\left(\dfrac{9}{4}-2x\right)\left(3x+\dfrac{7}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{9}{4}-2x=0\\3x+\dfrac{7}{8}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-2x=-\dfrac{9}{4}\\3x=-\dfrac{7}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{7}{24}\end{matrix}\right.\)