a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)

=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)

=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)

=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)

=\(1+0+14+1-\frac{19}{5}\)

=\(15+1-\frac{19}{5}\)

=\(16-\frac{19}{5}\)

=\(\frac{80}{5}-\frac{19}{5}\)

=\(\frac{61}{5}\)

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

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

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

\(\frac{x}{5}=\frac23\)

\(x\) = \(\frac23\times5\)

\(x=\frac{10}{3}\)

Vậy \(x=\frac{10}{3}\)

\(\frac{x}{3}-\frac12=\frac15\)

\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)

\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)

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

\(x=\frac{7}{10}\times3\)

\(x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\frac{x}{5}+\frac12=\frac{6}{10}\)

\(\frac{x}{5}=\frac{6}{10}-\frac12\)

\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)

\(\frac{x}{5}=\frac{1}{10}\)

\(x=\frac{1}{10}\times5\)

\(x=\frac12\)

Vậy \(x=\frac12\)

\(\frac{x+3}{15}\) = \(\frac13\)

\(x+3=\frac13\times15\)

\(x+3=5\)

\(x=5-3\)

\(x=2\)

Vậy \(x=2\)

a

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

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

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

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

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

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

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

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

a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)

(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)

\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)

\(\frac{44}{7}.x=\frac{-77}{35}\)

x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)

(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1

(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)

(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)

\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)

(44\(x\) + 3) : \(\frac{11}{5}\) = - 4

44\(x\) + 3 = - 4 x \(\frac{11}{5}\)

44\(x\) + 3 = - \(\frac{44}{5}\)

44\(x\) = - \(\frac{44}{5}\) - 3

44\(x\) = - \(\frac{59}{5}\)

\(x=-\frac{59}{5}:44\)

\(x\) = - \(\frac{59}{250}\)

Vậy \(x\) = - \(\frac{59}{250}\)

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, -5/6 -x = 7/12 + -1/3
⇔-10/12 - 12x/12 = 7/12 + -4/12
⇒-10 - 12x = 7 - 4
⇔-12x = 7 - 4 +10
⇔-12x = 13
⇔x = -13/12
b, x+13/-15 = 1/3
⇔-(x+13)/15 = 5/15
⇒ -x - 13 = 5
⇔-x = 5 +13
⇔-x = 18
⇔x = -18
c,-15/x-1 = -3/5
⇔-75/(x-1).5 = -3.(x-1)/5.(x-1)
⇒-75 = -3x + 3
⇔3x = 3 + 75
⇔3x = 78
⇔x = 26
d, (1/2).x + -2/5 = 1/5
⇔5x/10 + -4/10 = 1/10
⇒5x - 4 = 1
⇔5x = 1 + 4
⇔5x = 5
⇔x = 1
e, (-2/3).x + 1/5 = 1/10
⇔-20x/30 + 6/30 = 3/30
⇒-20x + 6 = 3
⇔-20x = 3 - 6
⇔-20x = -3
⇔x = 3/20
f, 4/5 - (1/2).x = 1/10
⇔8/10 - 5x/10 = 1/10
⇒8 - 5x = 1
⇔-5x = 1 - 8
⇔-5x = -7
⇔x=7/5