1) 3900:(10+x•15)=30 2)450:(25+5•x)=15 3)285-(x:14-481)=36•6 4)400-(18+705:x)•4=28•5

Let's solve each equation step by step:

1) \( 3900:(10+x\cdot15)=30 \)

This can be rewritten as:

\[ \frac{3900}{10 + x \cdot 15} = 30 \]

Cross multiplying gives:

\[ 3900 = 30(10 + 15x) \]

Expanding the right side:

\[ 3900 = 300 + 450x \]

Now, isolate \( x \):

\[ 3900 - 300 = 450x \] \[ 3600 = 450x \] \[ x = \frac{3600}{450} = 8 \]

2) \( 450:(25+5\cdot x)=15 \)

Rewriting it gives:

\[ \frac{450}{25 + 5x} = 15 \]

Cross multiplying leads to:

\[ 450 = 15(25 + 5x) \]

Expanding gives:

\[ 450 = 375 + 75x \]

Now simplify to find \( x \):

\[ 450 - 375 = 75x \] \[ 75 = 75x \] \[ x = 1 \]

3) \( 285 - (x:14 - 481) = 36 \cdot 6 \)

First, we calculate \( 36 \cdot 6 = 216 \):

Rewriting gives:

\[ 285 - \left(\frac{x}{14} - 481\right) = 216 \]

Rearranging gives:

\[ 285 - 216 = \frac{x}{14} - 481 \] \[ 69 = \frac{x}{14} - 481 \]

Add 481 to both sides:

\[ 69 + 481 = \frac{x}{14} \] \[ 550 = \frac{x}{14} \]

Multiplying both sides by 14 gives:

\[ x = 550 \cdot 14 = 7700 \]

4) \( 400 - (18 + 705:x) \cdot 4 = 28 \cdot 5 \)

Calculating \( 28 \cdot 5 = 140 \):

Rewriting gives:

\[ 400 - (18 + \frac{705}{x}) \cdot 4 = 140 \]

Rearranging gives:

\[ (18 + \frac{705}{x}) \cdot 4 = 400 - 140 \] \[ (18 + \frac{705}{x}) \cdot 4 = 260 \]

Dividing both sides by 4 gives:

\[ 18 + \frac{705}{x} = 65 \]

Subtracting 18 from both sides gives:

\[ \frac{705}{x} = 65 - 18 \] \[ \frac{705}{x} = 47 \]

Cross-multiplying gives:

\[ 705 = 47x \]

Finally, solving for \( x \):

\[ x = \frac{705}{47} = 15 \]

Summary of solutions:

1. \( x = 8 \)
2. \( x = 1 \)
3. \( x = 7700 \)
4. \( x = 15 \)