The expression you provided is:

$\frac{6}{5} \left( 1 - \frac{2}{7} \right) : \frac{7}{9} \left( \frac{81}{21} \right) + \frac{3}{4} \cdot \frac{6}{9}$

I'll go ahead and solve it step-by-step:

Step 1: Simplify the first term

$\frac{6}{5} \left( 1 - \frac{2}{7} \right)$ First, simplify inside the parentheses: $1 - \frac{2}{7} = \frac{7}{7} - \frac{2}{7} = \frac{5}{7}$ Now multiply: $\frac{6}{5} \times \frac{5}{7} = \frac{6 \times 5}{5 \times 7} = \frac{30}{35} = \frac{6}{7}$

Step 2: Simplify the second term

$\frac{7}{9} \times \frac{81}{21}$ First, simplify $\frac{81}{21}$: $\frac{81}{21} = \frac{81 \div 3}{21 \div 3} = \frac{27}{7}$ Now multiply: $\frac{7}{9} \times \frac{27}{7} = \frac{7 \times 27}{9 \times 7} = \frac{27}{9} = 3$

Step 3: Solve the division

The original expression had a division symbol between the first and second terms: $\frac{6}{7} : 3 = \frac{6}{7} \times \frac{1}{3} = \frac{6}{7 \times 3} = \frac{6}{21} = \frac{2}{7}$

Step 4: Simplify the third term

$\frac{3}{4} \times \frac{6}{9}$ Simplify $\frac{6}{9}$: $\frac{6}{9} = \frac{2}{3}$ Now multiply: $\frac{3}{4} \times \frac{2}{3} = \frac{3 \times 2}{4 \times 3} = \frac{6}{12} = \frac{1}{2}$

Step 5: Add the results

Now add the results of the division and the multiplication: $\frac{2}{7} + \frac{1}{2}$ To add these fractions, we need a common denominator. The least common denominator between 7 and 2 is 14: $\frac{2}{7} = \frac{4}{14}, \quad \frac{1}{2} = \frac{7}{14}$ Now add: $\frac{4}{14} + \frac{7}{14} = \frac{11}{14}$

Final Answer:

$\boxed{\frac{11}{14}}$

Would you like any more details or have any questions?

Here are 5 related questions you could explore:

1. How do you simplify complex fractions involving both division and multiplication?
2. What are the rules for dividing fractions?
3. How can you find a common denominator quickly when adding fractions?
4. How can the distributive property help in simplifying similar expressions?
5. How would you solve a more complex algebraic fraction with variables?

Tip: Always simplify fractions early in the process to make calculations easier.