Đáp án:

Bài 1 :

$a. (-\frac{1}{3})^{2} + \frac{5}{9} : ( \frac{1}{9} + \frac{-2}{3} )$

$= \frac{1}{9} + \frac{5}{9} : ( \frac{1}{9} - \frac{6}{9} )$

$= \frac{1}{9} + \frac{5}{9} : \frac{-5}{9}$

$= \frac{1}{9} - 1$

$= - \frac{8}{9}$

$b. 19,9.(-3,5) + 19,9.(-6,5)$

$= 19,9.( - 3,5 - 6,5 )$

$= 19,9.(-10)$

$= - 199$

$d. \frac{13}{17}.\frac{-7}{15} + \frac{8}{15}.\frac{-13}{17} + 2\frac{13}{17}$

$= \frac{-13}{17}.\frac{7}{15} + \frac{8}{15}.\frac{-13}{17} + 2 + \frac{13}{17}$

$= - \frac{13}{17}( \frac{7}{15} + \frac{8}{15} - 1 ) + 2$

$= - \frac{13}{17}( 1 - 1 ) + 2$

$= 2$

Bai 2 :

$a. x( - \frac{1}{15} )^{13} = (-\frac{1}{15} )^{15}$

⇔ $x = ( - \frac{1}{15} )^{2}$

⇔ $x = \frac{1}{225}$

$b. \frac{11}{12} + ( \frac{2}{5} - x ) = \frac{2}{3}$

⇔ $\frac{2}{5} - x = \frac{8}{12} - \frac{11}{12}$

⇔ $\frac{2}{5} - x = \frac{-3}{12}$

⇔ $\frac{2}{5} - x = - \frac{1}{4}$

⇔ $x = \frac{2}{5} + \frac{1}{4}$

⇔ $x = \frac{13}{20}$

$c. | 2x - \frac{1}{5} | = \frac{2}{5}$

⇔ $| \frac{10x-1}{5} | = \frac{2}{5}$

⇔ $| 10x - 1 | = 2$

⇔ \(\left[ \begin{array}{l}10x-1=2\\10x-1=-2\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}10x=3\\10x=-1\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{3}{10}\\x=-\frac{1}{10}\end{array} \right.\) 

Bai 3 :

$a. \frac{x}{-2,5} = \frac{4}{5}$

⇔ $x = \frac{4}{5}.(-2,5)$

⇔ $x = - 2$

$b. \frac{4}{7} : ( x - 1 ) = 2 : \frac{-7}{3}$ $( x \ne 1 )$

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

⇔ $\frac{4}{7} : ( x - 1 ) = \frac{-6}{7}$

⇔ $x - 1 = \frac{4}{7} : \frac{-6}{7}$

⇔ $x - 1 = \frac{-2}{3}$

⇔ $x = \frac{-2}{3} + 1$

⇔ $x = \frac{1}{3}$

$c. \frac{x+1}{2} = \frac{8}{x+1}$ $( x \ne - 1 )$

⇔ $( x + 1 )( x + 1 ) = 2.8$

⇔ $( x + 1 )^{2} = 16$

⇔ \(\left[ \begin{array}{l}x+1=4\\x+1=-4\end{array} \right.\) 

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