Đáp án: a) $x =\dfrac{5}{26}$
b)$x = \dfrac{-5,8}{3}=\dfrac{-29}{15}$
c)$x=\dfrac{5}{6}$
d)$x=-\dfrac{3}{20}$
e)\(\left[ \begin{array}{l}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{array} \right.\)
Giải thích các bước giải:
a) $\dfrac{2}{3}x + \dfrac{3}{2}x =\dfrac{5}{12}$
$<=> (\dfrac{2}{3} +\dfrac{3}{2})x =\dfrac{5}{12}$
$<=> (\dfrac{4}{6} +\dfrac{9}{6})x =\dfrac{5}{12}$
$<=> (\dfrac{4+9}{6})x =\dfrac{5}{12}$
$<=> \dfrac{13}{6}x =\dfrac{5}{12}$
$<=> x =\dfrac{5}{12} : \dfrac{13}{6}$
$<=> x =\dfrac{5.6}{12.13}$
$<=> x =\dfrac{30}{156}$
$<=> x =\dfrac{5}{26}$
.
b) $\dfrac{2}{5}+\dfrac{3}{5}. ( 3x-3,7) = -\dfrac{53}{10}$
$<=> \dfrac{3}{5}. ( 3x-3,7) = -\dfrac{53}{10}-\dfrac{2}{5}$
$<=> \dfrac{3}{5}. ( 3x-3,7) = -\dfrac{53}{10}-\dfrac{4}{10}$
$<=> \dfrac{3}{5}. ( 3x-3,7) = \dfrac{-53-4}{10}$
$<=> \dfrac{3}{5}. ( 3x-3,7) = \dfrac{-57}{10}$
$<=> 3x-3,7 = \dfrac{-57}{10} : \dfrac{3}{5}$
$<=> 3x-3,7 = \dfrac{-57.5}{10.3}$
$<=> 3x-3,7 = \dfrac{-285}{30}$
$<=> 3x-3,7 = -9,5$
$<=> 3x = -9,5+3,7$
$<=> 3x = -5,8$
$<=> x = -5,8 : 3$
$<=> x = \dfrac{-5,8}{3}=\dfrac{-29}{15}$
.
c) $\dfrac{7}{9} : ( 2 + \dfrac{3}{4} x) + \dfrac{5}{9} = \dfrac{23}{27}$
$<=> \dfrac{7}{9} : ( 2 + \dfrac{3}{4} x) = \dfrac{23}{27}-\dfrac{5}{9}$
$<=> \dfrac{7}{9} : ( 2 + \dfrac{3}{4} x) = \dfrac{23}{27}-\dfrac{15}{27}$
$<=> \dfrac{7}{9} : ( 2 + \dfrac{3}{4} x) = \dfrac{23-15}{27}= \dfrac{8}{27}$
$<=> 2 + \dfrac{3}{4} x = \dfrac{7}{9} : \dfrac{8}{27} $
$<=> 2 + \dfrac{3}{4} x = \dfrac{7.27}{9.8} = \dfrac{189}{72}= \dfrac{21}{8}$
$<=> \dfrac{3}{4} x = \dfrac{21}{8} - 2= \dfrac{21}{8} - \dfrac{16}{8}$
$<=> \dfrac{3}{4} x =\dfrac{21-16}{8} =\dfrac{5}{8}$
$<=> x = \dfrac{5}{8} : \dfrac{3}{4} = \dfrac{5.4}{8.3}=\dfrac{20}{24}=\dfrac{5}{6}$
.
d) $-\dfrac{2}{3}x + \dfrac{1}{5} = \dfrac{3}{10}$
$<=> -\dfrac{2}{3}x = \dfrac{3}{10}-\dfrac{1}{5}$
$<=> -\dfrac{2}{3}x = \dfrac{3}{10}-\dfrac{2}{10}=\dfrac{3-2}{10}=\dfrac{1}{10}$
$<=> -x =\dfrac{1}{10} : \dfrac{2}{3} = \dfrac{1.3}{10.2}=\dfrac{3}{20}$
$<=> x=-\dfrac{3}{20}$
.
$e) | x | - \dfrac{3}{4} =\dfrac{5}{3}$
$<=> | x | =\dfrac{5}{3} + \dfrac{3}{4}$
$<=> | x | =\dfrac{20}{12} + \dfrac{9}{12}=\dfrac{20+9}{12}=\dfrac{29}{12}$
$<=>$ \(\left[ \begin{array}{l}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{array} \right.\)
