$2)$

$a)$ 

$\dfrac{x}{-12} = \dfrac{3}{4}$

$\to \dfrac{x}{-12} =  \dfrac{9}{-12}$

$\to x =9$

Vậy $x =9$

$b)$

$( \dfrac{2}{5} - x)^3 . 50 \% = \dfrac{1}{2}$

$\to ( \dfrac{2}{5} - x)^3 . \dfrac{1}{2} = \dfrac{1}{2}$

$\to ( \dfrac{2}{5} - x)^3 = 1 = 1^3$

$ \to \dfrac{2}{5} - x = 1$

$\to x = \dfrac{2}{5} -1 =-  \dfrac{3}{5}$

Vậy $ x = -  \dfrac{3}{5}$

$c)$

$ 1 \dfrac{3}{2}x - \dfrac{2}{5}x = \dfrac{7}{15}$

$\to x ( 1 \dfrac{3}{2} - \dfrac{2}{5} ) = \dfrac{7}{15}$

$\to x ( \dfrac{5}{2} - \dfrac{2}{5} ) = \dfrac{7}{15}$

$\to x. \dfrac{21}{10} = \dfrac{7}{15}$

$\to x = \dfrac{2}{9}$

$d)$

$ |x- \dfrac{1}{5}|  - 0,75 = 4$

$ |x- \dfrac{1}{5}| = 4+0,75 = 4,75 = \dfrac{19}{4}$

TH1

$ x - \dfrac{1}{5} = \dfrac{19}{4}$

$ x = \dfrac{19}{4} + \dfrac{1}{5} = \dfrac{99}{20}$

TH2

$ x - \dfrac{1}{5} = -\dfrac{19}{4}$

$ x =  -\dfrac{19}{4} + \dfrac{1}{5} =-\dfrac{91}{20}$

Vậy $ x \in \{-\dfrac{91}{20} ; \dfrac{99}{20} \}$

Đáp án:

 2,4_3, -3_4,-11_6 , 0

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