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

\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = \(\dfrac{-3}{14}\) : \(\dfrac{5}{7}\) 

\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) =  - \(\dfrac{3}{10}\)

\(\dfrac{3}{5}\)\(x\)         = - \(\dfrac{3}{10}\) + \(\dfrac{11}{5}\)

\(\dfrac{3}{5}\)\(x\)       = \(\dfrac{19}{10}\) 

\(x\)         = \(\dfrac{19}{10}\) : \(\dfrac{3}{5}\)

\(x\)        = \(\dfrac{19}{6}\)

\(\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}:\dfrac{5}{7}\)

\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{10}\)

\(\Rightarrow\dfrac{3}{5}x=-\dfrac{3}{10}+\dfrac{11}{5}\)

\(\Rightarrow\dfrac{3}{5}x=\dfrac{19}{10}\)

\(\Rightarrow x=\dfrac{19}{10}:\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{19}{6}\)

a) x=4/2

b) x=1/12

c) 24/35

a,\(\dfrac{1}{2}+x=\dfrac{5}{2}\)

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

\(\Rightarrow x=2\)

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

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

\(\Rightarrow x=\dfrac{1}{12}\)

c,\(x:\dfrac{4}{5}=\dfrac{6}{7}\)

\(x=\dfrac{6}{7}.\dfrac{4}{5}\)

\(\Rightarrow\dfrac{24}{35}\)

a)1/2+x=5/2                                          b)3/4-x=2/3

    x=5/2-1/2=4/2=2                                  x=3/4-2/3=1/12

c)x:4/5=6/7

   x=6/7.4/5=24/35

\(\Rightarrow\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{3}\\ \Rightarrow a=5;b=3\)

\(=>\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\)

\(=>\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}=>\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}=\dfrac{5}{3}\)

vậy \(\dfrac{a}{b}=\dfrac{5}{3}\)

`(2 2/5 + 3/5 x) = 3/4`

`=> 12/5 + 3/5 x = 3/4`

`=> 3/5 x= 3/4 - 12/5`

`=> 3/5 x= -33/20`

`=> x= -33/20 : 3/5`

`=> x=-11/4`

Vậy `x= -11/4`

a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\) 

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

     \(\dfrac{x}{3}=\dfrac{-14}{15}\) 

\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)

\(x=\dfrac{-2}{7}+\dfrac{9}{7}\) 

\(x=1\)

x=1

Ta có: \(\dfrac{-5}{13}\cdot\dfrac{3}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)

\(=\dfrac{5}{13}\left(-\dfrac{3}{7}+\dfrac{1}{7}\right)-\dfrac{2}{17}\cdot\dfrac{8}{13}\)

\(=\dfrac{5}{13}\cdot\dfrac{-2}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}\)

\(=\dfrac{-10}{91}-\dfrac{16}{221}\)

\(=\dfrac{-282}{1547}\)

\(\dfrac{-5}{13}.\dfrac{3}{7}-\dfrac{2}{17}.\dfrac{8}{13}+\dfrac{5}{13}.\dfrac{1}{7}\) 

\(=\dfrac{5}{13}.\dfrac{-3}{7}-\dfrac{2}{17}.\dfrac{8}{13}+\dfrac{5}{13}.\dfrac{1}{7}\)

\(=\dfrac{5}{13}.\left(\dfrac{-3}{7}+\dfrac{1}{7}\right)+\dfrac{-2}{17}.\dfrac{8}{13}\) 

\(=\dfrac{5}{13}.\dfrac{-2}{7}+\dfrac{-2}{13}.\dfrac{8}{17}\)

\(=\dfrac{-2}{13}.\dfrac{5}{7}+\dfrac{-2}{13}.\dfrac{8}{17}\) 

\(=\dfrac{-2}{13}.\left(\dfrac{5}{7}+\dfrac{8}{17}\right)\) 

\(=\dfrac{-2}{13}.\dfrac{141}{119}\) 

\(=\dfrac{-282}{1547}\)

\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{5\times4\times13}{13\times5\times3}=\dfrac{4}{3}\\ b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\dfrac{3}{7}\times\dfrac{7}{5}+\dfrac{5}{2}\times\dfrac{7}{5}=\dfrac{3}{5}+\dfrac{7}{2}=\dfrac{6}{10}+\dfrac{35}{10}=\dfrac{41}{10}\\ c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)

a, 4/3  b, 41/10   c,22/45