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Đáp án:

 Bài 2:

`a)`

$\dfrac{2}{3}.x-\dfrac{1}{2}=$ `(\frac{-2}{3})^{2}`

$⇒\dfrac{2}{3}.x-\frac{1}{2}=\dfrac{4}{9}$

$⇒\dfrac{2}{3}.x=\dfrac{4}{9}+\dfrac{1}{2}$

$⇒\dfrac{2}{3}.x=\dfrac{8}{18}+\dfrac{9}{18}$

$⇒\dfrac{2}{3}.x=\dfrac{17}{18}$

$⇒x=\dfrac{17}{18}:\dfrac{2}{3}$

$⇒x=\dfrac{17}{18}.\dfrac{3}{2}$

$⇒ x=\dfrac{17.3}{6.3.2}$

$⇒x=\dfrac{17}{12}$

Vậy `x=\frac{17}{12}`

`b)`

`3,5-|2x-\frac{1}{2}|=-0,75`

`⇒ |2x-\frac{1}{2}|=3,5-(-0,75)`

`⇒ |2x-\frac{1}{2}|= 3,5+0,75`

`⇒ |2x-\frac{1}{2}|=4,25`

Xét 2 trường hợp:

 $+)TH1:$

$2x-\dfrac{1}{2}=4,25$

$⇒2x=4,25+\dfrac{1}{2}$

$⇒2x=4,75$

$⇒x=4,75:2$

$⇒x=\dfrac{19}{8}$

$+)TH2:$

$2x-\dfrac{1}{2}=-4,25$

$⇒2x=(-4,25)+\dfrac{1}{2}$

$⇒2x=(-3,75)$

$⇒x=(-3,75):2$

$⇒x=-\dfrac{15}{8}$

Vậy `x ∈ {\frac{19}{8};-\frac{15}{8}}`