Worksheet featuring five fraction-based order of operations problems with brackets, designed for practice in arithmetic and algebraic simplification.

Let's solve each of these problems step by step, following the order of operations (PEMDAS/BODMAS):
Parentheses, Exponents, Multiplication and Division (left to right), Addition and Subtraction (left to right).

We'll go through each expression one at a time.

---

1.

$$
(3 - \frac{1}{3})^2 - \frac{3}{5} \times \frac{1}{5} - 4^2
$$

Step 1: Simplify inside parentheses:
$$
3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}
$$

Step 2: Square it:
$$
\left(\frac{8}{3}\right)^2 = \frac{64}{9}
$$

Step 3: Compute multiplication:
$$
\frac{3}{5} \times \frac{1}{5} = \frac{3}{25}
$$

Step 4: Compute exponent:
$$
4^2 = 16
$$

Now substitute:
$$
\frac{64}{9} - \frac{3}{25} - 16
$$

Convert all to common denominator (LCM of 9, 25, 1 is 225):
- $\frac{64}{9} = \frac{64 \times 25}{225} = \frac{1600}{225}$
- $\frac{3}{25} = \frac{3 \times 9}{225} = \frac{27}{225}$
- $16 = \frac{3600}{225}$

Now compute:
$$
\frac{1600}{225} - \frac{27}{225} - \frac{3600}{225} = \frac{1600 - 27 - 3600}{225} = \frac{-2027}{225}
$$

✔ Answer: $-\frac{2027}{225}$

---

2.

$$
\left(\frac{2}{5} - \left(\frac{2}{3}\right)^2\right) \times \frac{1}{2} - \left(\frac{3}{2} - \frac{1}{2}\right)^2
$$

Step 1: Compute exponents:
$$
\left(\frac{2}{3}\right)^2 = \frac{4}{9}
$$

Step 2: Subtract inside first parentheses:
$$
\frac{2}{5} - \frac{4}{9} = \frac{18}{45} - \frac{20}{45} = -\frac{2}{45}
$$

Step 3: Multiply by $\frac{1}{2}$:
$$
-\frac{2}{45} \times \frac{1}{2} = -\frac{2}{90} = -\frac{1}{45}
$$

Step 4: Simplify second parentheses:
$$
\frac{3}{2} - \frac{1}{2} = 1 \Rightarrow 1^2 = 1
$$

Now subtract:
$$
-\frac{1}{45} - 1 = -\frac{1}{45} - \frac{45}{45} = -\frac{46}{45}
$$

✔ Answer: $-\frac{46}{45}$

---

3.

$$
(5 + \frac{1}{2})^2 - \frac{1}{3} \times 4^2 \times \frac{3}{4}
$$

Step 1: Add inside parentheses:
$$
5 + \frac{1}{2} = \frac{11}{2}
$$

Step 2: Square:
$$
\left(\frac{11}{2}\right)^2 = \frac{121}{4}
$$

Step 3: Compute $4^2 = 16$

Now compute multiplication:
$$
\frac{1}{3} \times 16 \times \frac{3}{4} = \frac{1}{3} \times \frac{3}{4} \times 16 = \left(\frac{1}{3} \cdot \frac{3}{4}\right) \cdot 16 = \frac{1}{4} \cdot 16 = 4
$$

Now subtract:
$$
\frac{121}{4} - 4 = \frac{121}{4} - \frac{16}{4} = \frac{105}{4}
$$

✔ Answer: $\frac{105}{4}$

---

4.

$$
\left(\frac{3}{2} - \frac{1}{2}\right)^2 + \frac{1}{2}\left(\frac{1}{3} - \left(\frac{3}{2}\right)^2\right)
$$

Step 1: First parentheses:
$$
\frac{3}{2} - \frac{1}{2} = 1 \Rightarrow 1^2 = 1
$$

Step 2: Compute $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$

Step 3: Inside second parentheses:
$$
\frac{1}{3} - \frac{9}{4} = \frac{4}{12} - \frac{27}{12} = -\frac{23}{12}
$$

Step 4: Multiply by $\frac{1}{2}$:
$$
\frac{1}{2} \times -\frac{23}{12} = -\frac{23}{24}
$$

Now add:
$$
1 + (-\frac{23}{24}) = \frac{24}{24} - \frac{23}{24} = \frac{1}{24}
$$

✔ Answer: $\frac{1}{24}$

---

5.

$$
\left(\frac{2}{3} + \left(\frac{3}{4}\right)^2\right) \times \frac{1}{2} + \left(\frac{1}{3} - \frac{1}{5}\right)^2
$$

Step 1: Compute $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$

Add:
$$
\frac{2}{3} + \frac{9}{16} = \frac{32}{48} + \frac{27}{48} = \frac{59}{48}
$$

Multiply by $\frac{1}{2}$:
$$
\frac{59}{48} \times \frac{1}{2} = \frac{59}{96}
$$

Step 2: $\frac{1}{3} - \frac{1}{5} = \frac{5}{15} - \frac{3}{15} = \frac{2}{15}$

Square:
$$
\left(\frac{2}{15}\right)^2 = \frac{4}{225}
$$

Now add:
$$
\frac{59}{96} + \frac{4}{225}
$$

Find LCM of 96 and 225.
Prime factors:
- 96 = $2^5 \times 3$
- 225 = $3^2 \times 5^2$
LCM = $2^5 \times 3^2 \times 5^2 = 32 \times 9 \times 25 = 7200$

Convert:
- $\frac{59}{96} = \frac{59 \times 75}{7200} = \frac{4425}{7200}$
- $\frac{4}{225} = \frac{4 \times 32}{7200} = \frac{128}{7200}$

Add:
$$
\frac{4425 + 128}{7200} = \frac{4553}{7200}
$$

✔ Answer: $\frac{4553}{7200}$

---

6.

$$
(3 + \frac{3}{2})^2 - \frac{3}{5} - 3^2 \times \frac{2}{5}
$$

Step 1: Add:
$$
3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2}
$$

Square:
$$
\left(\frac{9}{2}\right)^2 = \frac{81}{4}
$$

Step 2: $3^2 = 9$, then:
$$
9 \times \frac{2}{5} = \frac{18}{5}
$$

Now expression becomes:
$$
\frac{81}{4} - \frac{3}{5} - \frac{18}{5}
$$

Combine like terms:
$$
\frac{81}{4} - \left(\frac{3}{5} + \frac{18}{5}\right) = \frac{81}{4} - \frac{21}{5}
$$

LCM of 4 and 5 is 20:
- $\frac{81}{4} = \frac{405}{20}$
- $\frac{21}{5} = \frac{84}{20}$

Subtract:
$$
\frac{405 - 84}{20} = \frac{321}{20}
$$

✔ Answer: $\frac{321}{20}$

---

7.

$$
(5 + \frac{1}{4})^2 - \frac{1}{3} + \frac{3}{4} \times 3^2
$$

Step 1: $5 + \frac{1}{4} = \frac{21}{4}$

Square:
$$
\left(\frac{21}{4}\right)^2 = \frac{441}{16}
$$

Step 2: $3^2 = 9$

Then: $\frac{3}{4} \times 9 = \frac{27}{4}$

Now expression:
$$
\frac{441}{16} - \frac{1}{3} + \frac{27}{4}
$$

Convert all to denominator 48 (LCM of 16, 3, 4):

- $\frac{441}{16} = \frac{441 \times 3}{48} = \frac{1323}{48}$
- $\frac{1}{3} = \frac{16}{48}$
- $\frac{27}{4} = \frac{27 \times 12}{48} = \frac{324}{48}$

Now:
$$
\frac{1323}{48} - \frac{16}{48} + \frac{324}{48} = \frac{1323 - 16 + 324}{48} = \frac{1631}{48}
$$

✔ Answer: $\frac{1631}{48}$

---

8.

$$
\left((\frac{3}{4})^2 - \frac{1}{2}\right) \times \frac{3}{2} - \left(\frac{3}{2} + \frac{1}{2}\right)^2
$$

Step 1: $(\frac{3}{4})^2 = \frac{9}{16}$

Then:
$$
\frac{9}{16} - \frac{1}{2} = \frac{9}{16} - \frac{8}{16} = \frac{1}{16}
$$

Multiply by $\frac{3}{2}$:
$$
\frac{1}{16} \times \frac{3}{2} = \frac{3}{32}
$$

Step 2: $\frac{3}{2} + \frac{1}{2} = 2 \Rightarrow 2^2 = 4$

Now subtract:
$$
\frac{3}{32} - 4 = \frac{3}{32} - \frac{128}{32} = -\frac{125}{32}
$$

✔ Answer: $-\frac{125}{32}$

---

9.

$$
(\frac{1}{6} + \frac{1}{2})^2 + \frac{1}{3} \cdot \frac{3}{2} + (\frac{2}{5})^2
$$

Step 1: $\frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$

Square:
$$
\left(\frac{2}{3}\right)^2 = \frac{4}{9}
$$

Step 2: $\frac{1}{3} \cdot \frac{3}{2} = \frac{3}{6} = \frac{1}{2}$

Step 3: $(\frac{2}{5})^2 = \frac{4}{25}$

Now add:
$$
\frac{4}{9} + \frac{1}{2} + \frac{4}{25}
$$

Find LCM of 9, 2, 25 → LCM = $2 \times 3^2 \times 5^2 = 450$

Convert:
- $\frac{4}{9} = \frac{4 \times 50}{450} = \frac{200}{450}$
- $\frac{1}{2} = \frac{225}{450}$
- $\frac{4}{25} = \frac{4 \times 18}{450} = \frac{72}{450}$

Add:
$$
\frac{200 + 225 + 72}{450} = \frac{497}{450}
$$

✔ Answer: $\frac{497}{450}$

---

10.

$$
(\frac{1}{3} + \frac{1}{3})^2 - \frac{1}{5}(\frac{1}{5} + (\frac{3}{4})^2)
$$

Step 1: $\frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Square:
$$
\left(\frac{2}{3}\right)^2 = \frac{4}{9}
$$

Step 2: $(\frac{3}{4})^2 = \frac{9}{16}$

Then:
$$
\frac{1}{5} + \frac{9}{16} = \frac{16}{80} + \frac{45}{80} = \frac{61}{80}
$$

Now multiply by $\frac{1}{5}$:
$$
\frac{1}{5} \times \frac{61}{80} = \frac{61}{400}
$$

Now subtract:
$$
\frac{4}{9} - \frac{61}{400}
$$

LCM of 9 and 400 = 3600

- $\frac{4}{9} = \frac{4 \times 400}{3600} = \frac{1600}{3600}$
- $\frac{61}{400} = \frac{61 \times 9}{3600} = \frac{549}{3600}$

Subtract:
$$
\frac{1600 - 549}{3600} = \frac{1051}{3600}
$$

✔ Answer: $\frac{1051}{3600}$

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $-\frac{2027}{225}$ |
| 2 | $-\frac{46}{45}$ |
| 3 | $\frac{105}{4}$ |
| 4 | $\frac{1}{24}$ |
| 5 | $\frac{4553}{7200}$ |
| 6 | $\frac{321}{20}$ |
| 7 | $\frac{1631}{48}$ |
| 8 | $-\frac{125}{32}$ |
| 9 | $\frac{497}{450}$ |
| 10 | $\frac{1051}{3600}$ |

Let me know if you'd like them as mixed numbers or decimals!