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

               \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)

                 \(\dfrac{4}{5}\).\(x\) = 1

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

b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)

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

                \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)

                  \(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )

                  \(x\) = - \(\dfrac{20}{29}\)

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

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

      \(\dfrac{4}{7}\).\(x\)       = - \(\dfrac{13}{15}\)

           \(x\)     = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)

            \(x\)    = - \(\dfrac{91}{60}\)

d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)

     \(\dfrac{5}{7}\): \(x\)     = \(\dfrac{2}{3}\)+ 1

       \(\dfrac{5}{7}\): \(x\)   = \(\dfrac{5}{3}\)

              \(x\)  = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)

               \(x\) = \(\dfrac{3}{7}\)

a, 2/5 + 3/4 : x = -1/2

3/4 : x = -1/2 - 2/5

3/4 : x = -9/10

x = 3/4 : -9/10

x = -5/6

b, 5/7 - 2/3 . x = 4/5 

2/3 . x = 4/5 + 5/7

2/3 . x = 53/35

x = 53/35 : 2/3

x = 159/70

c và d mình làm dược nhưng ko ghi được cái suy ra

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a) x = 4/27 - 2/3

    x = -14/27

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

\(a,-\dfrac{3}{5}-x=-0,75\\ -\dfrac{3}{5}-x=-\dfrac{3}{4}\\ x=-\dfrac{3}{5}-\left(-\dfrac{3}{4}\right)\\ x=-\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{3}{20}\\ ---\\ b,1\dfrac{4}{5}=-0,15-x\\ \dfrac{9}{5}=-\dfrac{3}{20}-x\\ x=-\dfrac{3}{20}-\dfrac{9}{5}\\ x=-\dfrac{3}{20}-\dfrac{36}{20}\\ x=-\dfrac{39}{20}\\ ----\\ c,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}+\dfrac{4}{7}\\ \dfrac{33}{10}-x=\dfrac{26}{21}\\ x=\dfrac{33}{10}-\dfrac{26}{21}\\ x=\dfrac{433}{210}\)

a) 1/20 - (x - 8/5) = 1/10

x - 8/5 = 1/20 - 1/10

x - 8/5 = -1/20

x = -1/20 + 8/5

x = 31/20

b) 7/4 - (x + 5/3) = -12/5

x + 5/3 = 7/4 + 12/5

x + 5/3 = 83/20

x = 83/20 - 5/3

x = 149/60

c) x - [17/2 - (-3/7 + 5/3)] = -1/3

x - (17/2 - 26/21) = -1/3

x - 305/42 = -1/3

x = -1/3 + 305/42

x = 97/14

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

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

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

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

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

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

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

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

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

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

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

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

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

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

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

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

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

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

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

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

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }

\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)

   \(x=\dfrac{7}{35}+\dfrac{-15}{35}\)

   \(x=-\dfrac{8}{35}\)

\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}:\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}\times\dfrac{2}{3}\)

                 \(x=\dfrac{8}{21}\)

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

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

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

           \(x=-\dfrac{7}{6}-\dfrac{3}{4}\)

           \(x=-\dfrac{23}{12}\)

\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)

    \(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)

     \(\dfrac{-5}{9}-x=\dfrac{13}{18}\)

                \(x=\dfrac{-5}{9}-\dfrac{13}{18}\)

                \(x=\dfrac{-10}{18}-\dfrac{13}{18}\)

                \(x=-\dfrac{23}{18}\)

Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{138}{30}< X< \dfrac{200}{3}\)

\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)

Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)

\(\Rightarrow X=\dfrac{105381}{52}\)