c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)

vậy x = \(\frac{262}{35}\) 

d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\) 

vậy x = \(\frac{57}{8}\) 

e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)

vậy x = \(\frac{1}{4}\)

a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)

vậy x = \(\frac{35}{12}\)

b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)

vậy x = \(\frac{19}{14}\) 

a) \(x=\frac{7}{20}\)

b) \(x=\frac{7}{12}\)

c)\(x=\frac{8}{15}\)

a ) \(\frac{7}{8}:x=3-\frac{1}{2}\)

     \(\frac{7}{8}:x=\frac{5}{2}\)

            \(x=\frac{7}{8}:\frac{5}{2}\)

              \(x=0,35\)

b ) \(x+\frac{1}{2}.\frac{1}{3}=\frac{3}{4}\)

            \(x+\frac{1}{6}=\frac{3}{4}\)

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

                        \(x=\frac{7}{12}\)

c ) \(\frac{3}{2}.\frac{4}{5}-x=\frac{2}{3}\)

          \(\frac{7}{10}-x=\frac{2}{3}\)

                        \(x=\frac{7}{10}-\frac{2}{3}\)

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

d ) \(x.3\frac{1}{3}=3\frac{1}{3}:4\frac{1}{4}\)

      \(x:\frac{10}{3}=\frac{10}{3}:\frac{17}{4}\)

      \(x:\frac{10}{3}=\frac{40}{51}\)

                \(x=\frac{40}{51}:\frac{10}{3}\)

                \(x=\frac{4}{17}\)

e ) \(5\frac{2}{3}:x=3\frac{2}{3}-2\frac{1}{2}\)

      \(\frac{17}{3}:x=\frac{11}{3}-\frac{5}{2}\)

      \(\frac{17}{3}:x=\frac{7}{6}\)

                \(x=\frac{17}{3}:\frac{7}{6}\)

                \(x=\frac{34}{7}\)

Nếu mình đúng thì các bạn k mình nhé

a, x=\(\frac{7}{30}\)

Câu 1a:

75% - 1 1/2 + 0,5 : 5/12 - (-1/2)^2

= 0,75 - 1,5 + 0,5 x 12/5 - 0,25

= (0,75 - 0,25) - 1,5 + 1,2

= 0,5 - 1,5 + 1,2

= - 1 + 1,2

= 0,2

Bài 1b:

(5/7.0,6 - 5: 3 1/2). (40% - 1,4).(-2)^3

= (3/7 - 5 x 2/7).(0,4 - 1,4).8

= - 1.(-1).8

= 8

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)

        \(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)

        \(x-\frac{3}{5}=\frac{263}{210}\)

                    \(x=\frac{263}{210}+\frac{3}{5}\)

                    \(x=\frac{389}{210}\)

        VẬY: \(x=\frac{389}{210}\)

a) \(\frac{9}{20}\)                                                   c) \(\frac{-55}{4}\)                                                                                                                                 

b) \(\frac{116}{75}\)                                                d) \(\frac{-76}{45}\)

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