a

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

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

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

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

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

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

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

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

Bài 1:

a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)

\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)

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

b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)

\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)

\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)

\(=\frac{1}{2}-\frac{4}{5}\)

\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)

d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)

\(=5^2+2^5-\frac{15^2}{15^2}\)

\(=25+32-1\)

\(=56\)

e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)

\(=\frac{4}{17}+\frac{13}{17}\)

\(=\frac{17}{17}=1\)

g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)

\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)

\(=\frac{7}{12}\cdot4=\frac{7}{3}\)

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

Bài 2: 

a: =>x/7=1/21

=>x=1/3

c: =>x(3x-2)=0

=>x=0 hoặc x=2/3

Bài1:

a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)

b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)

c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)

a) \(x.\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)

\(\Leftrightarrow x.\left(-\frac{5}{6}\right)=\frac{5}{12}\)

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

b) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(\Leftrightarrow2+\frac{3}{4}x=\frac{21}{8}\)

\(\Leftrightarrow\frac{3}{4}x=\frac{5}{8}\)

\(\Leftrightarrow x=\frac{5}{6}\)

c) \(\frac{3}{5}.\left(3x-3,7\right)=-\frac{57}{10}\)

\(\Leftrightarrow3x-3,7=-\frac{19}{2}\)

\(\Leftrightarrow3x=-\frac{29}{5}\)

\(\Leftrightarrow x=-\frac{29}{15}\)

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

\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)

\(\Rightarrow-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}-\frac{1}{3}+1\)

\(\Rightarrow-7x=\frac{-1}{6}\)

\(\Rightarrow x=\frac{1}{42}\)

Vậy ...

\(\)

\(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Rightarrow3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{18}\)

Vậy...