Đáp án:
bài 1;
$a) (\dfrac{5}{9} - \dfrac{6}{11}) + (\dfrac{4}{9} - \dfrac{5}{11}+\dfrac{3}{5})$
$ = (\dfrac{5}{9} +\dfrac{4}{9}) +(-\dfrac{6}{11}-\dfrac{5}{11}) +\dfrac{3}{5}$
$ = 1+ (-1) + \dfrac{3}{5} = \dfrac{3}{5}$
$b)(\dfrac{4}{35} +\dfrac{5}{8})-(\dfrac{7}{9} - \dfrac{3}{8}-\dfrac{31}{35})$
$ = (\dfrac{4}{35} + \dfrac{31}{35}) + (\dfrac{5}{8} + \dfrac{3}{8}) - \dfrac{7}{9}$
$ = 1 + 1 - \dfrac{7}{9} = \dfrac{11}{9}$
$c)(\dfrac{25}{19} -\dfrac{13}{7}) -(\dfrac{9}{8} + \dfrac{6}{19}+\dfrac{1}{7})$
$ = (\dfrac{25}{19} -\dfrac{6}{19}) + (-\dfrac{13}{7} - \dfrac{1}{7}) -\dfrac{9}{8}$
$ = 1-2 - \dfrac{9}{8} = -\dfrac{17}{8}$
Bài 2 :
$a)\dfrac{2}{15}x+\dfrac{1}{3} =\dfrac{5}{6}x - \dfrac{1}{2}$
$ ⇔ \dfrac{2}{15}x- \dfrac{5}{6}x = -\dfrac{1}{2} - \dfrac{1}{3}$
$ ⇔-\dfrac{7}{10}x= -\dfrac{5}{6}$
$ ⇔ x = -\dfrac{5}{6} : (-\dfrac{7}{10})$
$⇔x = \dfrac{25}{21}$
$\text{Vậy x= $\dfrac{25}{21}$}$
$b)\dfrac{4}{3} - \dfrac{5}{6}x-\dfrac{1}{2} = -1 + \dfrac{3}{2}x$
$ ⇔ -\dfrac{5}{6}x - \dfrac{3}{2}x = -1 +\dfrac{1}{2} -\dfrac{4}{3}$
$⇔ -\dfrac{7}{3}x = -\dfrac{11}{6}$
$⇔x = -\dfrac{11}{6} : (-\dfrac{7}{3})$
$⇔ x= \dfrac{11}{14}$
$\text{Vậy x = $\dfrac{11}{14}$}$
$c)\dfrac{7}{9} - (\dfrac{1}{3}x-\dfrac{4}{9}) = \dfrac{5}{9} + \dfrac{2}{3}x$
$⇔\dfrac{7}{9} - \dfrac{1}{3}x +\dfrac{4}{9} =\dfrac{5}{9} +\dfrac{2}{3}x$
$⇔ -\dfrac{1}{3}x - \dfrac{2}{3}x= \dfrac{5}{9} -\dfrac{4}{9} -\dfrac{7}{9}$
$⇔ -x = -\dfrac{2}{3}$
$⇔x =\dfrac{2}{3}$
$\text{Vậy x = $\dfrac{2}{3}$}$
$d)\dfrac{9}{10} + \dfrac{3}{2}x =\dfrac{5}{6} - (\dfrac{2}{5} - \dfrac{7}{2}x)$
$⇔\dfrac{9}{10} + \dfrac{3}{2}x = \dfrac{5}{6} - \dfrac{2}{5} + \dfrac{7}{2}x$
$⇔ \dfrac{3}{2}x - \dfrac{7}{2}x = \dfrac{5}{6} - \dfrac{2}{5} - \dfrac{9}{10}$
$⇔-2x = -\dfrac{7}{15}$
$⇔x = -\dfrac{7}{15} : (-2)$
$⇔x = \dfrac{7}{30}$
$\text{Vậy x =$\dfrac{7}{30}$}$
