195. \( \sqrt{\left[\left(\frac{2}{5}-\frac{1}{15}\right)^{2}:\left(\frac{3}{10}+\frac{3}{2}\right)+\frac{1}{9}\right]+\frac{2}{81}}+\sqrt{\left[\left(\frac{9}{10} \times \frac{2}{3}+\frac{1}{5}\right)-\left(1-\frac{1}{2^{2}}\right)\right] \times \frac{10}{18}} \)
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Let's solve the expression step by step. First, we simplify the components inside the square roots starting with the first part: 1. Compute \( \frac{2}{5} - \frac{1}{15} \): - To subtract these fractions, we need a common denominator. The least common multiple of 5 and 15 is 15. - Convert \( \frac{2}{5} \) to have a denominator of 15: \( \frac{2}{5} = \frac{6}{15} \) - Now perform the subtraction: \( \frac{6}{15} - \frac{1}{15} = \frac{5}{15} = \frac{1}{3} \) 2. Now square this result: \( \left(\frac{1}{3}\right)^{2} = \frac{1}{9} \) 3. Compute \( \frac{3}{10} + \frac{3}{2} \): - The least common multiple of 10 and 2 is 10. - Convert \( \frac{3}{2} \) to have a denominator of 10: \( \frac{3}{2} = \frac{15}{10} \) - Now perform the addition: \( \frac{3}{10} + \frac{15}{10} = \frac{18}{10} = \frac{9}{5} \) 4. Now deal with the division: We have \( \frac{1}{9} : \frac{9}{5} \), which is equivalent to multiplying by the reciprocal: \( \frac{1}{9} \times \frac{5}{9} = \frac{5}{81} \) 5. Add \( \frac{1}{9} \) to \( \frac{5}{81} \): - Again, we need a common denominator. The least common multiple of 9 and 81 is 81. - Convert \( \frac{1}{9} \) to have a denominator of 81: \( \frac{1}{9} = \frac{9}{81} \) - Now perform the addition: \( \frac{5}{81} + \frac{9}{81} = \frac{14}{81} \) 6. Finally, add \( \frac{2}{81} \): \( \frac{14}{81} + \frac{2}{81} = \frac{16}{81} \) Now we have: \[ \sqrt{\frac{16}{81}} = \frac{4}{9} \] Now let's calculate the second component: 1. Compute \( \frac{9}{10} \times \frac{2}{3} \): \( \frac{9 \times 2}{10 \times 3} = \frac{18}{30} = \frac{3}{5} \) 2. Now compute \( 1 - \frac{1}{2^{2}} \): \( \frac{1}{2^{2}} = \frac{1}{4} \) and thus \( 1 - \frac{1}{4} = \frac{3}{4} \) 3. Now perform the subtraction: \( \frac{3}{5} - \frac{3}{4} \) - The least common multiple of 5 and 4 is 20. - Convert \( \frac{3}{5} \) to have a denominator of 20: \( \frac{3}{5} = \frac{12}{20} \) - Convert \( \frac{3}{4} \) to have a denominator of 20: \( \frac{3}{4} = \frac{15}{20} \) - Perform the subtraction: \( \frac{12}{20} - \frac{15}{20} = \frac{-3}{20} \) 4. Now multiply by \( \frac{10}{18} \): \( \frac{-3}{20} \times \frac{10}{18} = \frac{-30}{360} = -\frac{1}{12} \) 5. Now we need to take the square root. Since square root of a negative number is not real, we can write this as: \( \sqrt{-\frac{1}{12}} \) which indicates the presence of imaginary numbers. However, the context of the problem primarily seems to indicate this may not have been intended. Assuming we handle it in this context, efforts may often yield that solutions converge to either
