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Let's go through each of the given expressions step by step and simplify them to the simplest fractional form as requested.

### Expression D:
\[
D = 3 - \frac{20}{21} \times \frac{7}{5}
\]

1. **Multiply the fractions**:
\[
\frac{20}{21} \times \frac{7}{5} = \frac{20 \times 7}{21 \times 5} = \frac{140}{105}
\]
Now simplify \( \frac{140}{105} \). The greatest common divisor (GCD) of 140 and 105 is 35:
\[
\frac{140}{105} = \frac{140 \div 35}{105 \div 35} = \frac{4}{3}
\]

2. **Subtract from 3**:
\[
D = 3 - \frac{4}{3}
\]
Convert 3 into a fraction:
\[
3 = \frac{9}{3}
\]
Now subtract:
\[
D = \frac{9}{3} - \frac{4}{3} = \frac{9 - 4}{3} = \frac{5}{3}
\]

Thus, \( D = \frac{5}{3} \).

---

### Expression E:
\[
E = \left( \frac{2}{7} - \frac{1}{3} \right) \times \frac{7}{4} - \frac{5}{4}
\]

1. **Subtract the fractions inside the parentheses**:
\[
\frac{2}{7} - \frac{1}{3}
\]
To subtract, find a common denominator (LCM of 7 and 3 is 21):
\[
\frac{2}{7} = \frac{6}{21}, \quad \frac{1}{3} = \frac{7}{21}
\]
Now subtract:
\[
\frac{6}{21} - \frac{7}{21} = \frac{-1}{21}
\]

2. **Multiply by \( \frac{7}{4} \)**:
\[
\frac{-1}{21} \times \frac{7}{4} = \frac{-1 \times 7}{21 \times 4} = \frac{-7}{84}
\]
Simplify \( \frac{-7}{84} \):
\[
\frac{-7}{84} = \frac{-1}{12}
\]

3. **Now subtract \( \frac{5}{4} \)**:
Convert \( \frac{5}{4} \) to have a denominator of 12:
\[
\frac{5}{4} = \frac{15}{12}
\]
Now subtract:
\[
\frac{-1}{12} - \frac{15}{12} = \frac{-1 - 15}{12} = \frac{-16}{12}
\]
Simplify:
\[
\frac{-16}{12} = \frac{-4}{3}
\]

Thus, \( E = \frac{-4}{3} \).

---

### Expression F:
\[
F = \frac{1}{45} : \frac{4}{9} + \frac{7}{3}
\]

1. **First, divide \( \frac{1}{45} \) by \( \frac{4}{9} \)**:
Division of fractions is equivalent to multiplying by the reciprocal:
\[
\frac{1}{45} \div \frac{4}{9} = \frac{1}{45} \times \frac{9}{4} = \frac{1 \times 9}{45 \times 4} = \frac{9}{180}
\]
Simplify \( \frac{9}{180} \):
\[
\frac{9}{180} = \frac{1}{20}
\]

2. **Now add \( \frac{7}{3} \)**:
Convert \( \frac{7}{3} \) to have a denominator of 20:
\[
\frac{7}{3} = \frac{140}{60}, \quad \frac{1}{20} = \frac{3}{60}
\]
Now add:
\[
\frac{3}{60} + \frac{140}{60} = \frac{143}{60}
\]

Thus, \( F = \frac{143}{60} \).

---

### Expression G:
\[
G = \left( \frac{-5}{6} + \frac{3}{4} \right) : \left( \frac{7}{6} - \frac{6}{7} \right)
\]

1. **Simplify the numerator \( \frac{-5}{6} + \frac{3}{4} \)**:
Find the common denominator (LCM of 6 and 4 is 12):
\[
\frac{-5}{6} = \frac{-10}{12}, \quad \frac{3}{4} = \frac{9}{12}
\]
Now add:
\[
\frac{-10}{12} + \frac{9}{12} = \frac{-1}{12}
\]

2. **Simplify the denominator \( \frac{7}{6} - \frac{6}{7} \)**:
Find the common denominator (LCM of 6 and 7 is 42):
\[
\frac{7}{6} = \frac{49}{42}, \quad \frac{6}{7} = \frac{36}{42}
\]
Now subtract:
\[
\frac{49}{42} - \frac{36}{42} = \frac{13}{42}
\]

3. **Now divide**:
\[
\frac{\frac{-1}{12}}{\frac{13}{42}} = \frac{-1}{12} \times \frac{42}{13} = \frac{-1 \times 42}{12 \times 13} = \frac{-42}{156}
\]
Simplify \( \frac{-42}{156} \):
\[
\frac{-42}{156} = \frac{-7}{26}
\]

Thus, \( G = \frac{-7}{26} \).

---

### Final results:
- \( D = \frac{5}{3} \)
- \( E = \frac{-4}{3} \)
- \( F = \frac{143}{60} \)
- \( G = \frac{-7}{26} \)

Would you like further clarification on any of these calculations? Here are some related questions for more practice:

1. How do you find the least common multiple (LCM) of two numbers?
2. How do you simplify a complex fraction step by step?
3. What is the process for converting mixed numbers to improper fractions?
4. How do you multiply fractions and simplify them?
5. How can you approach a problem where multiple operations are combined (like in expression G)?

**Tip**: When dividing fractions, remember to multiply by the reciprocal to simplify the process efficiently.

Calculez chaque expression et donnez le résultat sous la forme d'une fraction la plus simple possible pour les expressions D, E, F, et G.

Learn how to simplify the fraction expressions D, E, F, and G with detailed step-by-step calculations. This exercise covers key concepts of multiplying, dividing, and adding fractions, including how to find the least common multiple (LCM). Suitable for students in Grades 8-10.

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