`# \text {04th5.}`
`(2x - 3) \times (3/4x + 1) = 0`
`=>` TH1: `2x - 3 = 0`
`=> 2x = 0 + 3`
`=> 2x = 3`
`=> x = 3/2`
TH2: `3/4x + 1 = 0`
`=> 3/4x = 0 - 1`
`=> 3/4x = -1`
`=> x = -1 \div 3/4`
`=> x = -3/4`
Vậy, `x \in {-3/4; 3/2}.`
\(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)