a) \(x+\frac{5}{12}=-1\frac{2}{7}\)

\(\Leftrightarrow x+\frac{5}{12}=\frac{-9}{7}\)

\(\Leftrightarrow x=\frac{-143}{84}\)

Vậy ...

b) \(4\frac{1}{2}x:\frac{5}{12}=0,5\)

\(\Leftrightarrow\frac{9}{2}x=\frac{5}{24}\)

\(\Leftrightarrow x=\frac{5}{108}\)

vậy...

c) \(7,5.1\frac{3}{4}x=6\frac{2}{5}\)

\(\Leftrightarrow\frac{105}{8}x=\frac{32}{5}\)

\(\Leftrightarrow x=\frac{256}{525}\)

Vậy ...

câu a sai đề nha

a, \(\frac{x}{5}=\frac{2}{3}\Leftrightarrow x=\frac{10}{3}\)

b, \(\frac{x}{-24}=\frac{20}{42}\Leftrightarrow x=-\frac{80}{7}\)

c, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow3x+9=15\Leftrightarrow x=2\)

d, \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\Leftrightarrow x\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\Leftrightarrow-\frac{5}{6}x=\frac{5}{12}\Leftrightarrow x=-\frac{1}{2}\)

a) 2 - ( \(5\frac{3}{8}\)x  X  - \(\frac{5}{24}\)) = \(\frac{5}{12}\)

             \(5\frac{3}{8}\)x  X  - \(\frac{5}{24}\)= \(\frac{19}{12}\)

             \(5\frac{3}{8}\)x  X             = \(\frac{43}{24}\)

                            X             = \(\frac{1}{3}\)

b)  \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x  X + \(\frac{1}{6}\))  = \(\frac{22}{23}\)

                    \(3\frac{1}{3}\)x  X + \(\frac{1}{6}\)   = \(\frac{23}{18}\)

                   \(3\frac{1}{3}\)x  X                = \(\frac{10}{9}\)

                                  X                 =\(\frac{1}{3}\)

C)  \(\frac{4}{5}\)x    X  - \(\frac{1}{2}\)x    X + \(\frac{3}{4}\)x    X = \(\frac{7}{40}\)

  (   \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\))  x   X                 = \(\frac{7}{40}\)

               \(\frac{21}{20}\)         x   X                      = \(\frac{7}{40}\)

                                       X                       =\(\frac{1}{6}\)

|\(\frac32x\) + \(\frac12\)| = |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 \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

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

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

đúng hết đấy nhé mình tính kĩ lắm ko sai đâu

       chúc may mắn

a) x = 1

a.X=\(\frac{25}{7}\)