1) \(\frac{1}{7}=\frac{8}{-x}\Rightarrow-x=7\cdot8\)

\(-x=56\)

\(x=-56\)

5) \(\left(x-2\frac{1}{4}\right):\left(-\frac{5}{6}\right)=3\)

\(x-\frac{9}{4}=3\cdot\left(-\frac{5}{6}\right)\)

\(x-\frac{9}{4}=-\frac{5}{2}\)

\(x=-\frac{5}{2}+\frac{9}{4}\)

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

6) \(-8\left(4\frac{1}{5}\cdot x+\frac{3}{10}\right)=4\frac{4}{9}\)

\(-8\left(\frac{21}{5}x+\frac{3}{10}\right)=\frac{40}{9}\)

\(\frac{21}{5}x+\frac{3}{10}=\frac{40}{9}:\left(-8\right)\)

\(\frac{21}{5}x+\frac{3}{10}=-\frac{5}{9}\)

\(\frac{21}{5}x=-\frac{5}{9}-\frac{3}{10}\)

\(\frac{21}{5}x=-\frac{77}{90}\)

\(x=-\frac{77}{90}:\frac{21}{5}\)

\(x=-\frac{11}{54}\)

1. \(x=\frac{61}{42}\)

2. \(x=\frac{-36}{5}\) 

3. \(x=\frac{13}{11}\)

4. \(x=\frac{1}{12}\)

5.\(x=\frac{-5}{2}\)

\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)

        \(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)

        \(x-\frac{3}{5}=\frac{263}{210}\)

                    \(x=\frac{263}{210}+\frac{3}{5}\)

                    \(x=\frac{389}{210}\)

        VẬY: \(x=\frac{389}{210}\)

\(x\times\frac{6}{25}=\frac{15}{-13}\)

x=\(\frac{15}{-13}\div\frac{6}{25}\)

x=\(-\frac{125}{26}\)

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\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)

\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)

\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)

\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)

\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)

|\(\frac32x\) + \(\frac12\)| = |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) \(\frac{-76}{45}\)

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