a) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)

=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)

=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-57}{10}\)

=> \(\left(3x-3,7\right)=\frac{-19}{5}\)

=>\(3x=\frac{-19}{5}+\frac{37}{10}\)

=>\(3x=\frac{-1}{10}\)

=>\(x=\frac{-1}{30}\)

Ko biết đúng hay sai!!?

4.x=29/12;-29/12

Câu cuối ban viết rõ lên nhé

a,x.3\(\dfrac{1}{4}\)+(\(\dfrac{-7}{6}\)).x-1\(\dfrac{2}{3}\)=\(\dfrac{5}{12}\)

\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x-\(\dfrac{5}{3}\)=\(\dfrac{5}{12}\)

\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x=\(\dfrac{25}{12}\)

\(\Leftrightarrow\)\(\dfrac{25}{12}\)x=\(\dfrac{25}{12}\)

\(\Leftrightarrow\)x=1

Vậy x=1

b,\(\dfrac{2}{5}\)+\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)

\(\Leftrightarrow\)\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)-\(\dfrac{2}{5}\)

\(\Leftrightarrow\)\(\dfrac{9}{5}\)x-\(\dfrac{111}{50}\)=\(\dfrac{-57}{10}\)

\(\Leftrightarrow\)\(\dfrac{9}{5}\)x=\(\dfrac{-87}{25}\)

\(\Leftrightarrow\)x=\(\dfrac{-29}{15}\)

Vậy x=\(\dfrac{-29}{15}\)

c,\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)+\(\dfrac{5}{9}\)=\(\dfrac{23}{27}\)

\(\Leftrightarrow\)\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)=\(\dfrac{8}{27}\)

\(\Leftrightarrow\)\(\dfrac{7}{18}\)+\(\dfrac{28}{27}\)x=\(\dfrac{8}{27}\)

\(\Leftrightarrow\)\(\dfrac{28}{27}\)x=\(\dfrac{-5}{54}\)

\(\Leftrightarrow\)x=\(\dfrac{-5}{56}\)

Vậy x=\(\dfrac{-5}{56}\)

a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)

    (\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)

     - \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)

     \(x\)   = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))

     \(x=\) - \(\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) 

b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)

           \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)

          \(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)

         3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)

         3\(x\)   - 3,7 = - \(\dfrac{19}{2}\)

         3\(x\)         = - \(\dfrac{19}{2}\) + 3,7

          3\(x\)        = - \(\dfrac{29}{5}\)

           \(x\)         = - \(\dfrac{29}{5}\) : 3

           \(x\)        = - \(\dfrac{29}{15}\)

Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\) 

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 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{21}{8}\)

             \(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2

             \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{5}{8}\)

               \(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)

              \(x\) =  \(\dfrac{5}{6}\)

Vậy \(x=\dfrac{5}{6}\)

a: =>|x+3/4|=2+1/5=11/5

=>x+3/4=11/5 hoặc x+3/4=-11/5

=>x=29/20 hoặc x=-59/20

b: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

c: =>|2x-1/3|=1/6

=>2x-1/3=1/6 hoặc 2x-1/3=-1/6

=>2x=1/2 hoặc 2x=1/6

=>x=1/4 hoặc x=1/12

e: =>x+2/3=0 hoặc -2x-3/5=0

=>x=-2/3 hoặc x=-3/10

a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)

\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)

\(x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\).

b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)

\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)

\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)

\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)

\(3x-3\cdot7=-\dfrac{19}{2}\)

\(3x-21=-\dfrac{19}{2}\)

\(3x=-\dfrac{19}{2}+21\)

\(3x=\dfrac{23}{2}\)

\(x=\)\(\dfrac{23}{2}:3\)

\(x=\dfrac{23}{6}\)

Vậy \(x=\dfrac{23}{6}\).

c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)

\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)

\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)

\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)

\(\dfrac{3}{4x}=-\dfrac{65}{22}\)

\(4x=\dfrac{3\cdot22}{-65}\)

\(4x=-\dfrac{66}{65}\)

\(x=-\dfrac{66}{65}:4\)

\(x=-\dfrac{33}{130}\)

Vậy \(x=-\dfrac{33}{130}\).

d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)

\(\left|x\right|=\dfrac{29}{12}\)

\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)

Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).