7. \( 6 \frac{25}{27}-4 \frac{4}{9}-1 \frac{1}{3} \)

To solve the expression \( 6 \frac{25}{27} - 4 \frac{4}{9} - 1 \frac{1}{3} \), first convert the mixed numbers to improper fractions: 1. \( 6 \frac{25}{27} = \frac{6 \times 27 + 25}{27} = \frac{162 + 25}{27} = \frac{187}{27} \) 2. \( 4 \frac{4}{9} = \frac{4 \times 9 + 4}{9} = \frac{36 + 4}{9} = \frac{40}{9} \) 3. \( 1 \frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{3 + 1}{3} = \frac{4}{3} \) Next, express all fractions with a common denominator. The least common multiple of 27, 9, and 3 is 27. Rewrite each term: 1. \( \frac{40}{9} = \frac{40 \times 3}{9 \times 3} = \frac{120}{27} \) 2. \( \frac{4}{3} = \frac{4 \times 9}{3 \times 9} = \frac{36}{27} \) Now substitute these into the expression: \[ \frac{187}{27} - \frac{120}{27} - \frac{36}{27} \] Combine the fractions: \[ \frac{187 - 120 - 36}{27} = \frac{31}{27} \] Since \( \frac{31}{27} \) is an improper fraction, it can be expressed as a mixed number: \[ 1 \frac{4}{27} \] Thus, the final answer is \( 1 \frac{4}{27} \).