$a$) `(x-1/3)(2x+4)=0`
`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x = -2\end{array} \right.\)
Vậy $x$ $∈$ `{1/3;-2}`
$b$) `(x-2/5)^2=9/25`
`⇔ (x-2/5)^2 = (± 3/5)^2`
`⇔ x - 2/5 = ± 3/5`
`⇒` \(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{5}\end{array} \right.\)
Vậy $x$ $∈$ `{1;-1/5}`
$c$) `(x-1/2)(2x+3)<0`
`⇒ x-1/2;2x+3` khác dấu
$TH1$. $\left\{\begin{matrix}x-1 > 0 & \\ 2x+ 3 < 0& \end{matrix}\right.$ $⇒$ $KTM$
$TH2$. $\left\{\begin{matrix}x-1 < 0 & \\ 2x+ 3 > 0& \end{matrix}\right.$ $⇒$ `-3/2 < x < 1`
Vậy `-3/2 < x < 1`
$d$) `5/{x-1} < 0`
`⇒ x-1 < 0`
`⇒ x < 1`
Vậy `x<1`
$e$) `{x+2}/{x-3} > 1`
`⇔ {x-3+5}/{x-3} > 1`
`⇔ 1 + 5/{x-3} > 1`
`⇒ 5/{x-3} > 0`
`⇔ x -3 > 0`
`⇔ x > 3`
Vậy `x>3`.
