Certainly! Let's solve for \(x\) in each equation from Bài 4 and Bài 5.

### Bài 4:

#### a) \( x + \frac{-1}{5} = \frac{-4}{15} \)

To solve for \( x \):

\[ x = \frac{-4}{15} + \frac{1}{5} \]

Convert \(\frac{1}{5}\) to have a common denominator of 15:

\[ \frac{1}{5} = \frac{3}{15} \]

Now add them:

\[ x = \frac{-4}{15} + \frac{3}{15} = \frac{-1}{15} \]

#### b) \( 3.7 - x = \frac{7}{10} \)

To solve for \( x \):

\[ x = 3.7 - \frac{7}{10} \]

Convert 3.7 to a fraction:

\[ 3.7 = \frac{37}{10} \]

Now subtract:

\[ x = \frac{37}{10} - \frac{7}{10} = \frac{30}{10} = 3 \]

#### c) \( x \cdot \frac{3}{2} = 2.4 \)

To solve for \( x \):

\[ x = \frac{2.4}{\frac{3}{2}} \]

Convert 2.4 to a fraction:

\[ 2.4 = \frac{24}{10} = \frac{12}{5} \]

Now divide by \(\frac{3}{2}\):

\[ x = \frac{12}{5} \times \frac{2}{3} = \frac{24}{15} = \frac{8}{5} \]

### Bài 5:

#### a) \( \frac{-2}{15} - x = \frac{-3}{10} \)

To solve for \( x \):

\[ x = \frac{-2}{15} + \frac{3}{10} \]

Convert \(\frac{3}{10}\) to have a common denominator of 30:

\[ \frac{3}{10} = \frac{9}{30} \quad \text{and} \quad \frac{-2}{15} = \frac{-4}{30} \]

Now add them:

\[ x = \frac{-4}{30} + \frac{9}{30} = \frac{5}{30} = \frac{1}{6} \]

#### b) \( x - \frac{1}{15} = \frac{1}{10} \)

To solve for \( x \):

\[ x = \frac{1}{10} + \frac{1}{15} \]

Find the common denominator (30):

\[ \frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30} \]

\[ x = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \]

#### c) \( \frac{-3}{8} - x = \frac{5}{12} \)

To solve for \( x \):

\[ x = \frac{-3}{8} - \frac{5}{12} \]

Convert to a common denominator (24):

\[ \frac{-3}{8} = \frac{-9}{24}, \quad \frac{5}{12} = \frac{10}{24} \]

\[ x = \frac{-9}{24} - \frac{10}{24} = \frac{-19}{24} \]

#### d) \( \frac{3}{5} \cdot x = \frac{-1}{4} + \frac{7}{10} \)

To solve for \( x \):

First, find the sum:

Convert to a common denominator (20):

\[ \frac{-1}{4} = \frac{-5}{20} \quad \text{and} \quad \frac{7}{10} = \frac{14}{20} \]

\[ \frac{-5}{20} + \frac{14}{20} = \frac{9}{20} \]

Now solve for \( x \):

\[ \frac{3}{5} \cdot x = \frac{9}{20} \]

\[ x = \frac{9}{20} \cdot \frac{5}{3} = \frac{45}{60} = \frac{3}{4} \]

#### e) \( \frac{-5}{8} - x = \frac{-3}{20} + \left(-\frac{1}{6}\right) \)

To solve for \( x \):

First, find the sum:

Convert to a common denominator (120):

\[ \frac{-3}{20} = \frac{-18}{120}, \quad -\frac{1}{6} = \frac{-20}{120} \]

\[ \frac{-18}{120} + \frac{-20}{120} = \frac{-38}{120} = \frac{-19}{60} \]

Now solve for \( x \):

\[ x = \frac{-5}{8} - \frac{-19}{60} \]

Convert \(\frac{-5}{8}\) to a common denominator (120):

\[ \frac{-5}{8} = \frac{-75}{120} \]

\[ x = \frac{-75}{120} + \frac{19}{60} = \frac{-75}{120} + \frac{38}{120} = \frac{-37}{120} \]

#### f) \( x - \left(-\frac{1}{4}\right) = -\frac{5}{6} + \frac{1}{8} \)

To solve for \( x \):

First, solve the right side:

Convert to a common denominator (24):

\[ -\frac{5}{6} = \frac{-20}{24}, \quad \frac{1}{8} = \frac{3}{24} \]

\[ -\frac{20}{24} + \frac{3}{24} = \frac{-17}{24} \]

Now solve for \( x \):

\[ x = \frac{-17}{24} - \frac{-1}{4} \]

Convert \(-\frac{1}{4}\) to a common denominator (24):

\[ -\frac{1}{4} = \frac{-6}{24} \]

\[ x = \frac{-17}{24} + \frac{6}{24} = \frac{-11}{24} \]

If you need further clarification, let me know!