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

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

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

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

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

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

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

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

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

|\(\frac32x\) + \(\frac12\)| = |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) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )

S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)

S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)

S = \(\frac12\) - \(\frac19\)

S = \(\frac{9}{18}-\frac{2}{18}\)

S = \(\frac{7}{18}\)

\(5,\left(x\cdot0,5-\frac{3}{7}\right):\frac{1}{2}=1\frac{1}{7}\)

\(\Leftrightarrow x\cdot0,5:\frac{1}{2}-\frac{3}{7}:\frac{1}{2}=1\frac{1}{7}\)

\(\Leftrightarrow x-\frac{6}{7}=\frac{8}{7}\)

\(\Leftrightarrow x=2\)

\(6,x\cdot1,75=1\frac{3}{10}+45\%\)

\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{13}{10}+\frac{9}{20}\)

\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{7}{4}\)

\(\Leftrightarrow x=1\)

\(7,\frac{5-x}{15}+\frac{5}{12}-\frac{1}{8}=\frac{3}{8}\)

\(\Leftrightarrow\frac{5-x}{15}=\frac{3}{8}-\frac{5}{12}+\frac{1}{8}\)

\(\Leftrightarrow\frac{5-x}{15}=\frac{1}{12}\)

\(\Leftrightarrow60-12x=15\)

\(\Leftrightarrow12x=45\)

\(\Leftrightarrow x=\frac{15}{4}\)

\(8,\left|x-\frac{25}{33}\right|-\frac{3}{11}=\frac{2}{3}\)

\(\Leftrightarrow\left|\frac{x-25}{33}\right|=\frac{31}{33}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{25}{33}=\frac{31}{33}\\x-\frac{25}{33}=-\frac{31}{33}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{56}{33}\\x=-\frac{2}{11}\end{cases}}\)

\(9,-\frac{9}{8}+\frac{-3}{8}\cdot x=-\frac{1}{8}\)

\(\Leftrightarrow\frac{-9}{8}+\frac{-3}{8}\cdot x+\frac{1}{8}=0\)

\(\Leftrightarrow-1-\frac{3}{8}x=0\)

\(\Leftrightarrow\frac{3}{8}x=-1\)

\(\Rightarrow x=-\frac{8}{3}\)

bạn có thể đừng làm tắt bước được không

Bài 1:

a) \(\frac{25}{4}+\frac{-5}{4}=\frac{25-5}{4}=\frac{20}{4}=5\)

b)\(\frac{-5}{9}+\left(\frac{-2}{7}\right)=\frac{-35}{63}+\left(\frac{-18}{63}\right)=\frac{-53}{63}\)

c) \(\frac{1}{4}+\frac{9}{11}+\frac{7}{4}+\left(\frac{-2}{11}\right)=\left(\frac{1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{11}+\frac{9}{11}\right)=2+\frac{7}{11}=\frac{22+7}{11}=\frac{29}{11}\)

d) \(\frac{-5}{19}.\frac{8}{19}+\left(\frac{-14}{19}\right).\frac{11}{19}=\frac{-40}{361}-\frac{151}{361}=-\frac{191}{361}\)

Bài 2: 

a) \(x+\frac{5}{9}=\frac{-8}{9}\)  \(\Leftrightarrow x=\frac{-8}{9}-\frac{5}{9}\)  \(\Leftrightarrow x=-\frac{13}{9}\)

b) \(\frac{-1}{8}-x=\frac{9}{20}\)  \(\Leftrightarrow x=\frac{-1}{8}-\frac{9}{20}\)  \(\Leftrightarrow x=\frac{-5}{40}-\frac{18}{40}\)  \(\Leftrightarrow x=-\frac{23}{40}\)

c) (x + 5)³ - 12 = 15

\(\Leftrightarrow\)(x + 5)³ = 27

\(\Leftrightarrow\)x + 5 = 3

\(\Leftrightarrow\)x = -2

d) \(\left|x-3\right|-\frac{4}{15}=\frac{26}{15}\)  \(\Leftrightarrow\left|x-3\right|=2\)  \(\Leftrightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)  \(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Bài 1:

\(g.\frac{5}{11}+\frac{6}{11}=\frac{5+6}{11}=\frac{11}{11}=1\)\(\)

\(e.\frac{-17}{25}.\frac{20}{33}+\frac{-17}{25}.\frac{13}{33}+\frac{-3}{25}=\frac{-17}{25}.\left(\frac{20}{33}+\frac{13}{33}\right)+\frac{-3}{25}\)

\(=\frac{-17}{25}.1+\frac{-3}{25}=\frac{-17}{25}+\frac{-3}{25}=\frac{-17-3}{25}=\frac{-20}{25}=\frac{-4}{5}\)

\(d.\frac{5}{7}.\frac{19}{23}+\frac{5}{7}.\frac{5}{23}-\frac{5}{7}.\frac{1}{23}=\frac{5}{7}\left(\frac{19}{23}+\frac{5}{23}-\frac{1}{23}\right)\)

\(=\frac{5}{7}\left(\frac{19+5-1}{23}\right)=\frac{5}{7}.1=\frac{5}{7}\)

\(c.\left(-11\right).\frac{9}{22}=\frac{\left(-11\right).9}{22}=\frac{-99}{22}=\frac{-9}{2}\)

\(b.\frac{5}{6}-\frac{1}{8}=\frac{5.4}{6.4}-\frac{1.3}{8.3}=\frac{20}{24}-\frac{3}{24}=\frac{17}{24}\)

\(a.\frac{2}{3}+\frac{1}{5}-\frac{1}{6}=\frac{2.10}{3.10}+\frac{1.6}{5.6}-\frac{1.5}{6.5}=\frac{20}{30}+\frac{6}{30}-\frac{5}{30}\)

\(=\frac{20+6-5}{30}=\frac{21}{30}=\frac{7}{10}\)

Bài 2:

\(a.\frac{3}{4}+x=\frac{11}{12}\)

\(\Leftrightarrow x=\frac{11}{12}-\frac{3}{4}\)

\(\Leftrightarrow x=\frac{11}{12}-\frac{9}{12}\)

\(\Leftrightarrow x=\frac{2}{12}=\frac{1}{6}\)

\(b.x-\frac{11}{12}=0,5\)

\(\Leftrightarrow x=\frac{1}{2}-\frac{11}{12}\)

\(\Leftrightarrow x=\frac{6}{12}+\frac{11}{12}\)

\(\Leftrightarrow x=\frac{17}{12}\)

0,7.2\(\frac23\).20.0,375.\(\frac{5}{28}\)

= \(\frac{7}{10}\) .\(\frac83\).20.\(\frac38\).\(\frac{5}{28}\)

= (\(\frac{7}{10}\).\(\frac{5}{28}\).20).(\(\frac83\).\(\frac38\))

= (\(\frac18.20\)). 1

= \(\frac52\)

B = (9,75 .21\(\frac37\) + \(\frac{39}{4}\).18\(\frac47\))

B = (\(\frac{39}{4}\).\(\frac{150}{7}\frac{}{7}\)+ \(\frac{39}{4}\).\(\frac{130}{7}\))

B = \(\frac{39}{4}\).(\(\frac{150}{7}\) + \(\frac{130}{7}\))

B = \(\frac{39}{4}\).\(\frac{280}{7}\)

B = 390