4. $\frac{x}{7}=\frac{3}{21}$ 9. $\frac{27}{m}=\frac{9}{2}$ 5. $\frac{3}{5}=\frac{6 x}{10}$ 10. $\frac{3 k}{5}=\frac{k+5}{10}$ 11. $\frac{3 x-4}{5}-\frac{2 x+1}{20}=\frac{-1}{2}-x$ 12. $\frac{x}{x-2}+\frac{2}{x+2}-\frac{x^{2}-1}{x^{2}-4}=0$

Let's solve each equation step by step:

---

### 4. $\frac{x}{7} = \frac{3}{21}$

First, simplify the right-hand side:

\[
\frac{3}{21} = \frac{1}{7}
\]

Now we have:

\[
\frac{x}{7} = \frac{1}{7}
\]

Since both sides have the same denominator, $x = 1$.

---

### 9. $\frac{27}{m} = \frac{9}{2}$

To solve for $m$, cross-multiply:

\[
27 \cdot 2 = 9 \cdot m
\]

This gives:

\[
54 = 9m
\]

Now, divide both sides by 9:

\[
m = \frac{54}{9} = 6
\]

---

### 5. $\frac{3}{5} = \frac{6x}{10}$

Simplify the right-hand side:

\[
\frac{6x}{10} = \frac{3x}{5}
\]

Now, since both sides have the same denominator:

\[
3 = 3x
\]

Divide both sides by 3:

\[
x = 1
\]

---

### 10. $\frac{3k}{5} = \frac{k+5}{10}$

Cross-multiply:

\[
3k \cdot 10 = 5 \cdot (k + 5)
\]

Simplify both sides:

\[
30k = 5(k + 5)
\]

Distribute on the right-hand side:

\[
30k = 5k + 25
\]

Move $5k$ to the left-hand side:

\[
30k - 5k = 25
\]

Simplify:

\[
25k = 25
\]

Divide both sides by 25:

\[
k = 1
\]

---

### 11. $\frac{3x - 4}{5} - \frac{2x + 1}{20} = \frac{-1}{2} - x$

To solve this, first find a common denominator on the left-hand side. The common denominator between 5 and 20 is 20, so rewrite both fractions:

\[
\frac{3x - 4}{5} = \frac{4(3x - 4)}{20} = \frac{12x - 16}{20}
\]

Now, the equation becomes:

\[
\frac{12x - 16}{20} - \frac{2x + 1}{20} = \frac{-1}{2} - x
\]

Combine the fractions on the left-hand side:

\[
\frac{(12x - 16) - (2x + 1)}{20} = \frac{-1}{2} - x
\]

Simplify the numerator:

\[
\frac{12x - 16 - 2x - 1}{20} = \frac{-1}{2} - x
\]

This becomes:

\[
\frac{10x - 17}{20} = \frac{-1}{2} - x
\]

Now, multiply through by 20 to eliminate the denominator:

\[
10x - 17 = 10\left(\frac{-1}{2} - x\right)
\]

Distribute on the right-hand side:

\[
10x - 17 = -5 - 10x
\]

Move all terms involving $x$ to one side:

\[
10x + 10x = -5 + 17
\]

Simplify:

\[
20x = 12
\]

Divide by 20:

\[
x = \frac{12}{20} = \frac{3}{5}
\]

---

### 12. $\frac{x}{x-2} + \frac{2}{x+2} - \frac{x^2 - 1}{x^2 - 4} = 0$

First, note that $x^2 - 4 = (x - 2)(x + 2)$, so rewrite the equation:

\[
\frac{x}{x - 2} + \frac{2}{x + 2} - \frac{x^2 - 1}{(x - 2)(x + 2)} = 0
\]

Now, express everything with a common denominator of $(x - 2)(x + 2)$:

\[
\frac{x(x + 2)}{(x - 2)(x + 2)} + \frac{2(x - 2)}{(x - 2)(x + 2)} - \frac{x^2 - 1}{(x - 2)(x + 2)} = 0
\]

Combine all the terms:

\[
\frac{x(x + 2) + 2(x - 2) - (x^2 - 1)}{(x - 2)(x + 2)} = 0
\]

Simplify the numerator:

\[
x(x + 2) = x^2 + 2x
\]
\[
2(x - 2) = 2x - 4
\]
\[
x^2 - 1 = x^2 - 1
\]

Now, the equation becomes:

\[
\frac{(x^2 + 2x) + (2x - 4) - (x^2 - 1)}{(x - 2)(x + 2)} = 0
\]

Simplify the numerator:

\[
x^2 + 2x + 2x - 4 - x^2 + 1 = 4x - 3
\]

Now, the equation is:

\[
\frac{4x - 3}{(x - 2)(x + 2)} = 0
\]

For the fraction to equal 0, the numerator must be 0:

\[
4x - 3 = 0
\]

Solve for $x$:

\[
4x = 3 \quad \Rightarrow \quad x = \frac{3}{4}
\]

---

Let me know if you would like any details about these steps!

### Related Questions:

1. How do you determine a common denominator in fraction equations?
2. What are the properties of cross-multiplication?
3. How can you simplify complex fractions?
4. What happens when you cross-multiply terms with variables?
5. How do you work with quadratic expressions in equations?

### Tip:
Always simplify fractions early to make the equation easier to solve later on.

4. $\frac{x}{7} = \frac{3}{21}$ 9. $\frac{27}{m} = \frac{9}{2}$ 5. $\frac{3}{5} = \frac{6x}{10}$ 10. $\frac{3k}{5} = \frac{k+5}{10}$ 11. $\frac{3x-4}{5} - \frac{2x+1}{20} = \frac{-1}{2} - x$ 12. $\frac{x}{x-2} + \frac{2}{x+2} - \frac{x^2 - 1}{x^2 - 4} = 0$

Learn how to solve rational equations, simplify complex fractions, and apply cross-multiplication techniques. This step-by-step guide covers key algebraic concepts and methods like finding common denominators and solving linear and quadratic expressions. Suitable for students in Grades 9-11.

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