`d,`
`(3 - 2x) . (x + 4) = 0`
`=>` \(\left[ \begin{array}{l}3 - 2x = 0\\x + 4 =0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x = \dfrac{3}{2}\\x=-4\end{array} \right.\)
Vậy `x ∈ {3/2 ; -4}`
`e,`
`x . x = 2x`
`=> x . x - 2x = 0`
`=> x^2 - 2x = 0`
`=> (x - 2) . (x + 0) = 0`
`=>` \(\left[ \begin{array}{l}x - 2 = 0\\x + 0 = 0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
Vậy `x ∈ {2 ; 0}`
`f,`
`(3x - 1)^2 = 16`
`=>` \(\left[ \begin{array}{l}(3x - 1)^2 = 4^2\\(3x - 1)^2 = (-4)^2\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}3x - 1 = 4\\3x - 1 = -4\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=-1\end{array} \right.\)
Vậy `x ∈ {5/3 ; -1}`
