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

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

   \(x=-\dfrac{8}{35}\)

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

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

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

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

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

                 \(x=\dfrac{8}{21}\)

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

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

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

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

           \(x=-\dfrac{23}{12}\)

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

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

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

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

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

                \(x=-\dfrac{23}{18}\)

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

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

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

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: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)

hay \(x=\dfrac{6}{25}\)

Vậy: \(x=\dfrac{6}{25}\)

b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)

hay \(x=\dfrac{5}{99}\)

Vậy: \(x=\dfrac{5}{99}\)

c) Ta có: \(\dfrac{-5}{9}+\dfrac{x}{10}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{x}{10}=\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{3}{9}+\dfrac{5}{9}=\dfrac{8}{9}\)

hay \(x=\dfrac{80}{9}\)

Vậy: \(x=\dfrac{80}{9}\)

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

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

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

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

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

b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)

=>x-1/2=27

hay x=55/2

c: =>1/2x-3/4=42/63=2/3

=>1/2x=17/12

hay x=17/6