To solve the expression \( \frac{3}{5} \times \frac{15}{39} \times 1 \frac{3}{5} \div 1 \frac{4}{5} \), let's first convert the mixed numbers into improper fractions.
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Convert \(1 \frac{3}{5}\): \[ 1 \frac{3}{5} = \frac{5 \times 1 + 3}{5} = \frac{5 + 3}{5} = \frac{8}{5} \]
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Convert \(1 \frac{4}{5}\): \[ 1 \frac{4}{5} = \frac{5 \times 1 + 4}{5} = \frac{5 + 4}{5} = \frac{9}{5} \]
Now, we substitute these into the expression: \[ \frac{3}{5} \times \frac{15}{39} \times \frac{8}{5} \div \frac{9}{5} \]
Next, we can rewrite the division as multiplication by the reciprocal: \[ \frac{3}{5} \times \frac{15}{39} \times \frac{8}{5} \times \frac{5}{9} \]
Now we can multiply the fractions together. First, let's simplify \( \frac{15}{39} \): \[ \frac{15}{39} = \frac{15 \div 3}{39 \div 3} = \frac{5}{13} \]
Now we can substitute that back into the expression: \[ \frac{3}{5} \times \frac{5}{13} \times \frac{8}{5} \times \frac{5}{9} \]
Now cancel out common factors:
- The \(5\) in the numerator and \(5\) in the denominator can cancel: \[ = \frac{3 \times 1 \times 8 \times 1}{1 \times 13 \times 1 \times 9} = \frac{3 \times 8}{13 \times 9} = \frac{24}{117} \]
Next, we can simplify \( \frac{24}{117} \): Find the greatest common divisor (gcd) of 24 and 117. The gcd is 3. \[ \frac{24 \div 3}{117 \div 3} = \frac{8}{39} \]
Thus, the final answer is: \[ \boxed{\frac{8}{39}} \]