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

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

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

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

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

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

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

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

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

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 = \(\frac{19}{14}\) 

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. 

| x | + \(\left|-\frac{2}{5}\right|=-\frac{5}{3}\)

| x | + \(\frac{2}{5}=-\frac{5}{3}\)

| x | = \(-\frac{5}{3}-\frac{2}{5}\)

| x | = \(-\frac{31}{15}\)

\(\Rightarrow x\in\varnothing\)vì trị đối \(\ge\)0

Vậy x \(\in\varnothing\)

b.

| x - 3 | = \(\frac{4}{5}\)

\(\Rightarrow\)x - 3 = \(\frac{4}{5}\)hoặc \(-\frac{4}{5}\)

\(\Rightarrow\)x = \(\frac{4}{5}+3\)hoặc \(-\frac{4}{5}+3\)

\(\Rightarrow\)x = \(\frac{19}{5}\)hoặc \(\frac{11}{5}\)

Vậy x \(\in\){ \(\frac{19}{5}\); \(\frac{11}{5}\)} 

c.

| x - 7 | = \(\frac{5}{3}\)

\(\Rightarrow\)x - 7 = \(\frac{5}{3}\)hoặc \(-\frac{5}{3}\)

\(\Rightarrow\)x = \(\frac{5}{3}+7\)hoặc \(-\frac{5}{3}+7\)

\(\Rightarrow\)x = \(\frac{26}{3}\)hoặc \(\frac{16}{3}\)

Vậy x \(\in\){ \(\frac{26}{3}\); \(\frac{16}{3}\)}

d.

| x - \(\frac{1}{2}\)| = \(\frac{1}{4}\)

\(\Rightarrow\)x - \(\frac{1}{2}\)= \(\frac{1}{4}\)hoặc \(-\frac{1}{4}\)

\(\Rightarrow\)x = \(\frac{1}{4}+\frac{1}{2}\)hoặc \(-\frac{1}{4}+\frac{1}{2}\)

\(\Rightarrow\)x = \(\frac{3}{4}\)hoặc \(\frac{1}{4}\)

Vậy x \(\in\){ \(\frac{3}{4}\); \(\frac{1}{4}\)}

e.

| x - 7 | = \(-\frac{5}{3}\)

\(\Rightarrow\)x - 7 = \(-\frac{5}{3}\)hoặc \(\frac{5}{3}\)

\(\Rightarrow\)x = \(-\frac{5}{3}+7\)hoặc \(\frac{5}{3}+7\)

\(\Rightarrow\)x = \(\frac{16}{3}\)hoặc \(\frac{26}{3}\)

Vậy x \(\in\){ \(\frac{16}{3}\); \(\frac{26}{3}\)}

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) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)

b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)

c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)

d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)

e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)

Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((