Bài 2: 

b: x+25%x=-1,25

=>1,25x=-1,25

hay x=-1

c: x-75%x=1/4

=>1/4x=1/4

hay x=1

Bài 2: 

a: =3/2-11/4=6/4-11/4=-5/4

b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3

\(a.-1,52+\frac{2}{47}-x=3.\)

\(\Leftrightarrow\frac{2}{47}-x=4,52\)

\(\Leftrightarrow x=\frac{2}{47}-4,52\)

\(\Leftrightarrow x=-\frac{5261}{1175}\)

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

\(\Leftrightarrow x=\frac{3}{2}-\frac{8}{5}-\frac{7}{2}+\frac{4}{7}\)

\(\Leftrightarrow x=-\frac{106}{35}\)

\(c.\left(x:2\frac{1}{8}\right).\frac{1}{6}=-\frac{3}{8}.\left(\frac{1}{2}-1\frac{3}{5}\right)\)

\(\Leftrightarrow\left(x:\frac{17}{8}\right).\frac{1}{6}=-\frac{3}{16}+\frac{3}{5}\)

\(\Leftrightarrow\left(x:\frac{17}{8}\right).\frac{1}{6}=\frac{33}{80}\)

\(\Leftrightarrow x:\frac{17}{8}=\frac{99}{40}\)

\(\Leftrightarrow x=\frac{99}{40}.\frac{17}{8}\)

\(\Leftrightarrow x=\frac{1683}{320}\)

|\(\frac32x\) + \(\frac12\)| = |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

g)=>x+1/2=0

x=0-1/2

x=-1/2

hoặc 2/3-2x=0

2x=2/3-0

2x=2/3

x=2/3:2

x=1/3

nhìn @_@ hoa cả mắt đăng từng bài thôi bạn

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

a

\(5\frac{4}{7}:x+=13\)

\(\frac{39}{7}:x=13\)

\(x=\frac{39}{7}:13\)

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

\(\frac{4}{7}x=\frac{9}{8}-0,125\)

\(\frac{4}{7}x=1\)

\(x=1:\frac{4}{7}\)

\(x=\frac{7}{4}=1\frac{3}{4}\)