Set of 24 complex fraction problems for algebra practice.

Let's solve each of these 24 algebraic expressions step by step. These problems involve simplifying complex fractions, often using algebraic manipulation such as factoring, finding common denominators, and simplifying exponents (especially negative exponents). We'll go through each one systematically.

---

1.

$$
\frac{1 + \frac{2}{5}}{2 + \frac{3}{5}}
$$

Simplify numerator and denominator:
- Numerator: $1 + \frac{2}{5} = \frac{5}{5} + \frac{2}{5} = \frac{7}{5}$
- Denominator: $2 + \frac{3}{5} = \frac{10}{5} + \frac{3}{5} = \frac{13}{5}$

Now divide:
$$
\frac{\frac{7}{5}}{\frac{13}{5}} = \frac{7}{5} \cdot \frac{5}{13} = \frac{7}{13}
$$

✔ Answer: $\boxed{\frac{7}{13}}$

---

2.

$$
\frac{\frac{4}{x - 1}}{x - 1}
$$

This is:
$$
\frac{4}{x - 1} \div (x - 1) = \frac{4}{(x - 1)(x - 1)} = \frac{4}{(x - 1)^2}
$$

✔ Answer: $\boxed{\frac{4}{(x - 1)^2}}$

---

3.

$$
\frac{1 - \frac{2}{x}}{\frac{4}{x} + \frac{9}{x}}
$$

Wait — the denominator is:
$$
\frac{4}{x} + \frac{9}{x} = \frac{13}{x}
$$

Numerator:
$$
1 - \frac{2}{x} = \frac{x - 2}{x}
$$

So:
$$
\frac{\frac{x - 2}{x}}{\frac{13}{x}} = \frac{x - 2}{x} \cdot \frac{x}{13} = \frac{x - 2}{13}
$$

✔ Answer: $\boxed{\frac{x - 2}{13}}$

---

4.

$$
\frac{\frac{10}{3x}}{\frac{5}{6x}}
$$

Divide:
$$
\frac{10}{3x} \div \frac{5}{6x} = \frac{10}{3x} \cdot \frac{6x}{5} = \frac{10 \cdot 6x}{3x \cdot 5} = \frac{60x}{15x} = 4
$$

✔ Answer: $\boxed{4}$

---

5.

$$
\frac{4x^2 - y^2}{\frac{2}{y} - \frac{1}{x}}
$$

Note: $4x^2 - y^2 = (2x - y)(2x + y)$

Denominator: $\frac{2}{y} - \frac{1}{x} = \frac{2x - y}{xy}$

So:
$$
\frac{(2x - y)(2x + y)}{\frac{2x - y}{xy}} = (2x - y)(2x + y) \cdot \frac{xy}{2x - y} = (2x + y) \cdot xy
$$

Cancel $2x - y$:

$$
= xy(2x + y)
$$

✔ Answer: $\boxed{xy(2x + y)}$

---

6.

$$
\frac{\frac{x + 1}{3}}{\frac{2x - 1}{6}}
$$

Divide:
$$
\frac{x + 1}{3} \div \frac{2x - 1}{6} = \frac{x + 1}{3} \cdot \frac{6}{2x - 1} = \frac{6(x + 1)}{3(2x - 1)} = \frac{2(x + 1)}{2x - 1}
$$

✔ Answer: $\boxed{\frac{2(x + 1)}{2x - 1}}$

---

7.

$$
\frac{\frac{2}{x} + \frac{3}{x^2}}{\frac{4}{x^2} - \frac{9}{x}}
$$

Simplify numerator and denominator separately.

Numerator:
$$
\frac{2}{x} + \frac{3}{x^2} = \frac{2x + 3}{x^2}
$$

Denominator:
$$
\frac{4}{x^2} - \frac{9}{x} = \frac{4 - 9x}{x^2}
$$

Now divide:
$$
\frac{\frac{2x + 3}{x^2}}{\frac{4 - 9x}{x^2}} = \frac{2x + 3}{x^2} \cdot \frac{x^2}{4 - 9x} = \frac{2x + 3}{4 - 9x}
$$

Note: $4 - 9x = -(9x - 4)$, so:
$$
= -\frac{2x + 3}{9x - 4}
$$

✔ Answer: $\boxed{-\frac{2x + 3}{9x - 4}}$

---

8.

$$
\frac{\frac{1}{x} + \frac{2}{x^2}}{x + \frac{8}{x^2}}
$$

Numerator:
$$
\frac{1}{x} + \frac{2}{x^2} = \frac{x + 2}{x^2}
$$

Denominator:
$$
x + \frac{8}{x^2} = \frac{x^3 + 8}{x^2}
$$

Now divide:
$$
\frac{\frac{x + 2}{x^2}}{\frac{x^3 + 8}{x^2}} = \frac{x + 2}{x^2} \cdot \frac{x^2}{x^3 + 8} = \frac{x + 2}{x^3 + 8}
$$

But $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

So:
$$
\frac{x + 2}{(x + 2)(x^2 - 2x + 4)} = \frac{1}{x^2 - 2x + 4}, \quad x \ne -2
$$

✔ Answer: $\boxed{\frac{1}{x^2 - 2x + 4}}$

---

9.

$$
\frac{\frac{2}{x} + 3}{\frac{4}{x^2} - 9}
$$

Numerator:
$$
\frac{2}{x} + 3 = \frac{2 + 3x}{x}
$$

Denominator:
$$
\frac{4}{x^2} - 9 = \frac{4 - 9x^2}{x^2}
$$

Now:
$$
\frac{\frac{2 + 3x}{x}}{\frac{4 - 9x^2}{x^2}} = \frac{2 + 3x}{x} \cdot \frac{x^2}{4 - 9x^2} = \frac{(2 + 3x)x}{4 - 9x^2}
$$

Note: $4 - 9x^2 = -(9x^2 - 4) = -(3x - 2)(3x + 2)$

And $2 + 3x = 3x + 2$

So:
$$
= \frac{(3x + 2)x}{-(3x - 2)(3x + 2)} = \frac{x}{-(3x - 2)} = -\frac{x}{3x - 2}
$$

✔ Answer: $\boxed{-\frac{x}{3x - 2}}$

---

10.

$$
\frac{1 - \frac{x}{y}}{\frac{x^2}{y^2} - 1}
$$

Numerator:
$$
1 - \frac{x}{y} = \frac{y - x}{y}
$$

Denominator:
$$
\frac{x^2}{y^2} - 1 = \frac{x^2 - y^2}{y^2} = \frac{(x - y)(x + y)}{y^2}
$$

Now:
$$
\frac{\frac{y - x}{y}}{\frac{(x - y)(x + y)}{y^2}} = \frac{y - x}{y} \cdot \frac{y^2}{(x - y)(x + y)}
$$

Note: $y - x = -(x - y)$, so:
$$
= \frac{-(x - y)}{y} \cdot \frac{y^2}{(x - y)(x + y)} = -\frac{1}{y} \cdot \frac{y^2}{x + y} = -\frac{y}{x + y}
$$

✔ Answer: $\boxed{-\frac{y}{x + y}}$

---

11.

$$
\frac{\frac{-2x}{x^2 - xy}}{\frac{y}{x^2}}
$$

First simplify numerator:
$$
\frac{-2x}{x(x - y)} = \frac{-2}{x - y}
$$

Now divide:
$$
\frac{-2}{x - y} \div \frac{y}{x^2} = \frac{-2}{x - y} \cdot \frac{x^2}{y} = \frac{-2x^2}{y(x - y)}
$$

✔ Answer: $\boxed{\frac{-2x^2}{y(x - y)}}$

---

12.

$$
\frac{\frac{7y}{x^2 + xy}}{\frac{y^2}{x^2}}
$$

Numerator:
$$
\frac{7y}{x(x + y)}
$$

Denominator:
$$
\frac{y^2}{x^2}
$$

So:
$$
\frac{7y}{x(x + y)} \div \frac{y^2}{x^2} = \frac{7y}{x(x + y)} \cdot \frac{x^2}{y^2} = \frac{7y \cdot x^2}{x(x + y) \cdot y^2} = \frac{7x}{(x + y)y}
$$

✔ Answer: $\boxed{\frac{7x}{y(x + y)}}$

---

13.

$$
\frac{\frac{2}{x} + \frac{1}{x^2}}{\frac{y}{x^2} + 1}
$$

Numerator:
$$
\frac{2x + 1}{x^2}
$$

Denominator:
$$
\frac{y + x^2}{x^2}
$$

So:
$$
\frac{\frac{2x + 1}{x^2}}{\frac{x^2 + y}{x^2}} = \frac{2x + 1}{x^2 + y}
$$

✔ Answer: $\boxed{\frac{2x + 1}{x^2 + y}}$

---

14.

$$
\frac{\frac{5}{x^2} - \frac{2}{x}}{\frac{1}{x} + 2}
$$

Numerator:
$$
\frac{5 - 2x}{x^2}
$$

Denominator:
$$
\frac{1 + 2x}{x}
$$

Now:
$$
\frac{5 - 2x}{x^2} \div \frac{1 + 2x}{x} = \frac{5 - 2x}{x^2} \cdot \frac{x}{1 + 2x} = \frac{5 - 2x}{x(1 + 2x)}
$$

Note: $5 - 2x = -(2x - 5)$, but no further simplification.

✔ Answer: $\boxed{\frac{5 - 2x}{x(1 + 2x)}}$

---

15.

$$
\frac{\frac{x}{9} - \frac{1}{x}}{1 + \frac{3}{x}}
$$

Numerator:
$$
\frac{x^2 - 9}{9x}
$$

Denominator:
$$
\frac{x + 3}{x}
$$

Now:
$$
\frac{\frac{x^2 - 9}{9x}}{\frac{x + 3}{x}} = \frac{x^2 - 9}{9x} \cdot \frac{x}{x + 3} = \frac{(x - 3)(x + 3)}{9x} \cdot \frac{x}{x + 3} = \frac{x - 3}{9}
$$

✔ Answer: $\boxed{\frac{x - 3}{9}}$

---

16.

$$
\frac{\frac{x}{4} - \frac{4}{x}}{1 - \frac{4}{x}}
$$

Numerator:
$$
\frac{x^2 - 16}{4x}
$$

Denominator:
$$
\frac{x - 4}{x}
$$

Now:
$$
\frac{\frac{x^2 - 16}{4x}}{\frac{x - 4}{x}} = \frac{x^2 - 16}{4x} \cdot \frac{x}{x - 4} = \frac{(x - 4)(x + 4)}{4x} \cdot \frac{x}{x - 4} = \frac{x + 4}{4}
$$

✔ Answer: $\boxed{\frac{x + 4}{4}}$

---

17.

$$
\frac{x^{-1}}{x^{-2} + y^{-2}}
$$

Rewrite with positive exponents:
$$
\frac{\frac{1}{x}}{\frac{1}{x^2} + \frac{1}{y^2}} = \frac{1/x}{\frac{y^2 + x^2}{x^2 y^2}} = \frac{1}{x} \cdot \frac{x^2 y^2}{x^2 + y^2} = \frac{x y^2}{x^2 + y^2}
$$

✔ Answer: $\boxed{\frac{x y^2}{x^2 + y^2}}$

---

18.

$$
\frac{2a^{-1} + 3b^{-2}}{a^{-1} - b^{-1}}
$$

Rewrite:
$$
\frac{\frac{2}{a} + \frac{3}{b^2}}{\frac{1}{a} - \frac{1}{b}} = \frac{\frac{2b^2 + 3a}{ab^2}}{\frac{b - a}{ab}} = \frac{2b^2 + 3a}{ab^2} \cdot \frac{ab}{b - a} = \frac{2b^2 + 3a}{b(b - a)}
$$

✔ Answer: $\boxed{\frac{2b^2 + 3a}{b(b - a)}}$

---

19.

$$
\frac{x^{-1} + y^{-1}}{3x^{-2} + 5y^{-2}} = \frac{\frac{1}{x} + \frac{1}{y}}{\frac{3}{x^2} + \frac{5}{y^2}}
$$

Numerator:
$$
\frac{x + y}{xy}
$$

Denominator:
$$
\frac{3y^2 + 5x^2}{x^2 y^2}
$$

Now:
$$
\frac{x + y}{xy} \div \frac{3y^2 + 5x^2}{x^2 y^2} = \frac{x + y}{xy} \cdot \frac{x^2 y^2}{3y^2 + 5x^2} = \frac{(x + y) x y}{3y^2 + 5x^2}
$$

✔ Answer: $\boxed{\frac{xy(x + y)}{5x^2 + 3y^2}}$

---

20.

$$
\frac{x^{-2}}{x + 3x^{-1}} = \frac{\frac{1}{x^2}}{x + \frac{3}{x}} = \frac{1/x^2}{\frac{x^2 + 3}{x}} = \frac{1}{x^2} \cdot \frac{x}{x^2 + 3} = \frac{1}{x(x^2 + 3)}
$$

✔ Answer: $\boxed{\frac{1}{x(x^2 + 3)}}$

---

21.

$$
\frac{a^{-1} + 1}{a^{-1} - 1} = \frac{\frac{1}{a} + 1}{\frac{1}{a} - 1} = \frac{\frac{1 + a}{a}}{\frac{1 - a}{a}} = \frac{1 + a}{1 - a}
$$

✔ Answer: $\boxed{\frac{1 + a}{1 - a}}$

---

22.

$$
\frac{3x^{-1} + (2y)^{-1}}{x^{-2}} = \frac{\frac{3}{x} + \frac{1}{2y}}{\frac{1}{x^2}} = \left(\frac{3}{x} + \frac{1}{2y}\right) \cdot x^2 = 3x + \frac{x^2}{2y}
$$

✔ Answer: $\boxed{3x + \frac{x^2}{2y}}$

---

23.

$$
\frac{2a^{-1} + (2a)^{-1}}{a^{-1} + 2a^{-1}} = \frac{\frac{2}{a} + \frac{1}{2a}}{\frac{1}{a} + \frac{2}{a}} = \frac{\frac{4 + 1}{2a}}{\frac{3}{a}} = \frac{\frac{5}{2a}}{\frac{3}{a}} = \frac{5}{2a} \cdot \frac{a}{3} = \frac{5}{6}
$$

✔ Answer: $\boxed{\frac{5}{6}}$

---

24.

$$
\frac{5x^{-1} + 2y^{-1}}{x^{-2} y^{-2}} = \frac{\frac{5}{x} + \frac{2}{y}}{\frac{1}{x^2 y^2}} = \left(\frac{5}{x} + \frac{2}{y}\right) \cdot x^2 y^2 = 5x y^2 + 2x^2 y
$$

✔ Answer: $\boxed{5xy^2 + 2x^2 y}$

---

## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $\frac{7}{13}$ |
| 2 | $\frac{4}{(x - 1)^2}$ |
| 3 | $\frac{x - 2}{13}$ |
| 4 | $4$ |
| 5 | $xy(2x + y)$ |
| 6 | $\frac{2(x + 1)}{2x - 1}$ |
| 7 | $-\frac{2x + 3}{9x - 4}$ |
| 8 | $\frac{1}{x^2 - 2x + 4}$ |
| 9 | $-\frac{x}{3x - 2}$ |
| 10 | $-\frac{y}{x + y}$ |
| 11 | $\frac{-2x^2}{y(x - y)}$ |
| 12 | $\frac{7x}{y(x + y)}$ |
| 13 | $\frac{2x + 1}{x^2 + y}$ |
| 14 | $\frac{5 - 2x}{x(1 + 2x)}$ |
| 15 | $\frac{x - 3}{9}$ |
| 16 | $\frac{x + 4}{4}$ |
| 17 | $\frac{x y^2}{x^2 + y^2}$ |
| 18 | $\frac{2b^2 + 3a}{b(b - a)}$ |
| 19 | $\frac{xy(x + y)}{5x^2 + 3y^2}$ |
| 20 | $\frac{1}{x(x^2 + 3)}$ |
| 21 | $\frac{1 + a}{1 - a}$ |
| 22 | $3x + \frac{x^2}{2y}$ |
| 23 | $\frac{5}{6}$ |
| 24 | $5xy^2 + 2x^2 y$ |

Let me know if you'd like any explanation expanded!