Tìm x, biếta) \(x + \frac{1}{3} = \frac{1}{{12}}\)b) \(\frac{1}{5}.x - \frac{1}{{20}} =  - \frac{1}{4}\)

* Hướng dẫn giải

a) \(x + \frac{1}{3} = \frac{1}{{12}}\)

\(\begin{array}{l}
x = \frac{1}{{12}} - \frac{1}{3}\\
x = \frac{{ - 1}}{4}
\end{array}\)

b) \(\frac{1}{5}.x - \frac{1}{{20}} =  - \frac{1}{4}\)

\(\begin{array}{l}
\frac{1}{5}.x =  - \frac{1}{4} + \frac{1}{{20}}\\
\frac{1}{5}.x =  - \frac{1}{5}\\
x =  - \frac{1}{5}:\frac{1}{5}\\
x =  - 1
\end{array}\)