Đáp án:

$a. x = \frac{-105}{64}$

$b.$ \(\left[ \begin{array}{l}x=\frac{13}{10}\\x=-\frac{7}{10}\end{array} \right.\) 

$c. x = \frac{11}{8}$

$d. x = - \frac{49}{10}$

$e.$ \(\left[ \begin{array}{l}x=-\frac{13}{15}\\x=\frac{17}{15}\end{array} \right.\) 

$f. x = \frac{325}{2}$

Giải thích các bước giải:

$a. \frac{8x}{5} - \frac{4x}{3} = (-\frac{5}{16}).1\frac{2}{5}$

⇔ $\frac{24x}{15} - \frac{20x}{15} = \frac{-5}{16}.\frac{7}{5}$

⇔ $\frac{4x}{15} = - \frac{7}{16}$

⇔ $x = \frac{-7}{16} : \frac{4}{15}$

⇔ $x = \frac{-7}{16}.\frac{15}{4}$

⇔ $x = \frac{-105}{64}$

$b. ( \frac{2x}{3} - \frac{1}{5} )^{2} = \frac{4}{9}$

⇔ \(\left[ \begin{array}{l}\frac{2x}{3}-\frac{1}{5}=\frac{2}{3}\\\frac{2x}{3}-\frac{1}{5}=-\frac{2}{3}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{2x}{3}=\frac{2}{3}+\frac{1}{5}\\\frac{2x}{3}=-\frac{2}{3}+\frac{1}{5}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{2x}{3}=\frac{13}{15}\\\frac{2x}{3}=-\frac{7}{15}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{13}{15}:\frac{2}{3}\\x=-\frac{7}{15}:\frac{2}{3}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{13}{10}\\x=-\frac{7}{10}\end{array} \right.\) 

$c. ( \frac{1}{2} - \frac{4x}{7} )^{3} = - \frac{8}{343}$

⇔ $( \frac{1}{2} - \frac{4x}{7} )^{3} = (\frac{-2}{7})^{3}$

⇔ $\frac{1}{2} - \frac{4x}{7} = \frac{-2}{7}$

⇔ $\frac{4x}{7} = \frac{1}{2} + \frac{2}{7}$

⇔ $\frac{4x}{7} = \frac{7}{14} + \frac{4}{14}$

⇔ $\frac{4x}{7} = \frac{11}{14}$

⇔ $x = \frac{11}{14} : \frac{4}{7}$

⇔ $x = \frac{11}{14}.\frac{7}{4}$

⇔ $x = \frac{11}{8}$

$d. \frac{2x}{3} - \frac{x}{2} = (-\frac{7}{12}).1\frac{2}{5}$

⇔ $\frac{4x}{6} - \frac{3x}{6} = \frac{-7}{12}.\frac{7}{5}$

⇔ $\frac{x}{6} = \frac{-49}{60}$

⇔ $x = 6.\frac{-49}{60}$

⇔ $x = - \frac{49}{10}$

$e. ( \frac{1}{5} - \frac{3x}{2} )^{2} = \frac{9}{4}$

⇔ \(\left[ \begin{array}{l}\frac{1}{5}-\frac{3x}{2}=\frac{3}{2}\\\frac{1}{5}-\frac{3x}{2}=-\frac{3}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{3x}{2}=\frac{1}{5}-\frac{3}{2}\\\frac{3x}{2}=\frac{1}{5}+\frac{3}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{3x}{2}=\frac{-13}{10}\\\frac{3x}{2}=\frac{17}{10}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{-13}{10}:\frac{3}{2}\\x=\frac{17}{10}:\frac{3}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-\frac{13}{15}\\x=\frac{17}{15}\end{array} \right.\) 

$f. ( 125 - \frac{4x}{5} )^{3} = - 125$

⇔ $( 125 - \frac{4x}{5} )^{3} = (-5)^{3}$

⇔ $125 - \frac{4x}{5} = - 5$

⇔ $\frac{4x}{5} = 125 + 5$

⇔ $\frac{4x}{5} = 130$

⇔ $x = 130 : \frac{4}{5}$

⇔ $x = 130.\frac{5}{4}$

⇔ $x = \frac{325}{2}$