Lời giải:
`1)`
`(x+3/5)(3/2-2x)=0`
$⇔\left[\begin{matrix}x+\dfrac{3}{5}=0\\ \dfrac{3}{2}-2x=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{-3}{5}\\ -2x=\dfrac{-3}{2}\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{-3}{5}\\ x=\dfrac{3}{4}\end{matrix}\right.$
Vậy `x∈{-3/5;3/4}`
`2)`
`(-5/4.x+1/2)[3/5-(-2/3.x)]=0`
`⇔(-5/4.x+1/2)(3/5+2/3x)=0`
$⇔\left[\begin{matrix}\dfrac{-5}{4}x+\dfrac{1}{2}=0\\ \dfrac{3}{5}+\dfrac{2}{3}x=0\end{matrix}\right.$
$⇔\left[\begin{matrix}\dfrac{-5}{4}x=\dfrac{-1}{2}\\ \dfrac{2}{3}x=\dfrac{-3}{5}\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{2}{5}\\ x=\dfrac{-9}{10}\end{matrix}\right.$
Vậy `x∈{2/5;-9/10}`
