\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

mình giải từng bài nhá

hả đơn giản

4.x=16

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

Bài 1:

a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)

     \(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)

     \(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)

     \(x=-\) \(\dfrac{49}{56}\)

Vậy \(x=-\dfrac{49}{56}\)

b; 6 - \(x\) = - \(\dfrac{3}{4}\)

         \(x\) = 6 + \(\dfrac{3}{4}\)

         \(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)

         \(x=\dfrac{27}{4}\)

Vậy \(x=\dfrac{27}{4}\) 

c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)

              \(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)

              \(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)

               \(x=\dfrac{19}{20}\)

Vậy \(x=\dfrac{19}{20}\) 

      Bài 1:

d; - 6 - \(x\) = - \(\dfrac{3}{5}\)

      \(x\)   = - 6 + \(\dfrac{3}{5}\)

       \(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)

       \(x=-\dfrac{27}{5}\)

Vậy \(x=-\dfrac{27}{5}\)

e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)

             \(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)

             \(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)

              \(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)

               \(x=\dfrac{22}{21}\)

Vậy \(x=\dfrac{22}{21}\) 

f; - 8 - \(x\) =  - \(\dfrac{5}{3}\)

          \(x\) = \(-\dfrac{5}{3}\) + 8

         \(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)

         \(x\) = \(\dfrac{-19}{3}\)

Vậy \(x=-\dfrac{19}{3}\) 

Bài 1g; - \(\dfrac{2}{10}\) - \(x\) = - \(\dfrac{9}{8}\)

                \(x\) = - \(\dfrac{2}{10}\) + \(\dfrac{9}{8}\)

                \(x\) = - \(\dfrac{8}{40}\) + \(\dfrac{45}{40}\)

                 \(x\) = \(\dfrac{37}{40}\) 

               Vậy \(x=\dfrac{37}{40}\)

h; 8 - \(x\) = \(-\dfrac{1}{6}\)

           \(x\) = 8 + \(\dfrac{1}{6}\)

          \(x\) = \(\dfrac{48}{6}\) + \(\dfrac{1}{6}\)

          \(x=\dfrac{49}{6}\)

Vậy \(x=\dfrac{49}{6}\)

i; - \(\dfrac{2}{5}\) + \(x\) = \(\dfrac{5}{7}\)

          \(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{5}\)

          \(x\) = \(\dfrac{39}{35}\) 

Vậy \(x=\dfrac{39}{35}\)

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)

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

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

=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )

Vậy x=\(\varnothing\)

+) \(\dfrac{1}{3}x=-\dfrac{4}{3}-\dfrac{1}{2}=-\dfrac{11}{6}\)

\(x=-\dfrac{11}{6}:\dfrac{1}{3}=-\dfrac{11}{2}\)

+) \(\dfrac{4}{3}x=-\dfrac{2}{3}+\dfrac{1}{2}=-\dfrac{1}{6}\)

\(x=-\dfrac{1}{6}:\dfrac{4}{3}=-\dfrac{1}{8}\)

+) \(2\left(x-1\right)=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{19}{6}\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{31}{12}\)

\(\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{4}{3}\)

\(\dfrac{1}{3}x=\left(-\dfrac{4}{3}\right)-\dfrac{1}{2}\)

\(\dfrac{1}{3}x=-\dfrac{11}{6}\)

\(x=\left(-\dfrac{11}{6}\right):\dfrac{1}{3}\)

\(x=-\dfrac{11}{2}\)

\(-\dfrac{2}{3}-\dfrac{4}{3}x=-\dfrac{1}{2}\)

\(\dfrac{4}{3}x=\left(-\dfrac{2}{3}\right)-\dfrac{-1}{2}\)

\(\dfrac{4}{3}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\dfrac{4}{3}\)

\(x=-\dfrac{1}{8}\)

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

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

\(2\left(x-1\right)=\dfrac{19}{6}\)

\(\left(x-1\right)=\dfrac{19}{6}:2\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{19}{12}+1\)

\(x=\dfrac{31}{12}\)

\(\dfrac{7}{-2}x-\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\dfrac{-7}{2}x-\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\dfrac{-7}{2}x=\left(-\dfrac{1}{2}\right)+\dfrac{1}{3}\)

\(\dfrac{-7}{2}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\left(-\dfrac{7}{2}\right)\)

\(x=\dfrac{1}{21}\)

\(\dfrac{8}{5}-\dfrac{1}{2}:x=\dfrac{8}{3}\)

\(\dfrac{1}{2}:x=\dfrac{8}{5}-\dfrac{8}{3}\)

\(\dfrac{1}{2}:x=-\dfrac{16}{15}\)

\(x=\dfrac{1}{2}:\left(-\dfrac{16}{15}\right)\)

\(x=-\dfrac{15}{32}\)

\(-\dfrac{5}{4}x-\dfrac{1}{2}x=-\dfrac{7}{3}\)

\(x\cdot\left(-\dfrac{5}{4}-\dfrac{1}{2}\right)=-\dfrac{7}{3}\)

\(x\cdot\left(-\dfrac{7}{4}\right)=-\dfrac{7}{3}\)

\(x=\left(-\dfrac{7}{3}\right):\left(-\dfrac{7}{4}\right)\)

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