$1$.
`9/2-[2/3-(x+7/4)]={-5}/4`
`⇔ 2/3 - (x+7/4)= {23}/4`
`⇔ x + 7/4 = {-61}/12`
`⇔ x = {-82}/12`
`⇔ x = {-41}/6`
Vậy `⇔ x = {-41}/6`
$2$.
`(x-1/3).(x+2/3)=0`
$⇒$ \(\left[ \begin{array}{l}x-\dfrac{1}{3}=0\\x+\dfrac{2}{3}=0\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{array} \right.\)
Vậy $x$ $∈$ `{\frac{1}{3};-\frac{2}{3}}`
$3$.
`(3/{4}.x-9/{16}).(1/5+{-3}/{5}:x)=0`
$⇒$ \(\left[ \begin{array}{l}\dfrac{3}{4}.x-\dfrac{9}{16}=0\\\dfrac{1}{5}+\dfrac{-3}{5}:x=0\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}\dfrac{3}{4}.x=\dfrac{9}{16}\\\dfrac{-3}{5}:x=\dfrac{-1}{5}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{3}{4}\\x=3\end{array} \right.\)
Vậy $x$ $∈$ `{\frac{3}{4};3}`
