Giải thích các bước giải:

$a)1\dfrac{1}{5}-(\dfrac{1}{2}x+\dfrac{3}{4})=\dfrac{-2}{5}\\\dfrac{8}{5}=\dfrac{x}{2}+\dfrac{3}{4}\\\dfrac{17}{20}=\dfrac{x}{2}\\x=\dfrac{17}{10}\\$

$b)(4x-5)(\dfrac{5x}{4}-2)=0$

$TH1:4x-5=0$

$x=\dfrac{5}{4}$

$TH2: \dfrac{5x}{4}-2=0$

$x=\dfrac{8}{5}$

$c)\dfrac{16}{x+2}=\dfrac{x+2}{4}\\(x+2)^2=64$

$TH1:x+2=8$

$x=6$

$TH2:x+2=-8$

$x=-10$

$d)\dfrac{24}{27}-(3x-\dfrac{7}{9})^3=1\dfrac{5}{27}\\-\dfrac{8}{27}=(3x-\dfrac{7}{9})^3\\3x-\dfrac{7}{9}=\dfrac{-2}{3}\\3x=\dfrac{1}{9}\\x=\dfrac{1}{27}$