Question

61/3-[(17/6÷34/3)+(9÷9/4)]*4​

Answer 1

Answer:

$\frac{61}{3} - (( \frac{17}{6} \div \frac{34}{3} ) + (9 \div \frac{9}{4} )) \times 4 \\ \\ = \frac{61}{3} - (( \frac{17}{6} \times \frac{3}{34} ) + ( \frac{9}{1} \times \frac{4}{9} )) \times 4 \\ \\ = \frac{61}{3} - ( \frac{1}{4} + \frac{4}{1} ) \times 4 \\ \\ = \frac{61}{3} - ( \frac{16 + 1}{4} ) \times 4 \\ \\ = \frac{61}{3} - \frac{17}{4} \times \frac{4}{1} \\ \\ = \frac{61}{3} - \frac{17}{1} \\ \\ = \frac{61 - 51}{3} \\ \\ = \frac{10}{3} \\ \\ = 3 \frac{1}{3}$

Answer 2

Answer:

$\frac{61}{3} - (( \frac{17}{6} \div \frac{34}{3} ) + (9 \div \frac{9}{4} )) \times 4 \\ \\ = \frac{61}{3} - (( \frac{17}{6} \times \frac{3}{34} ) + ( \frac{9}{1} \times \frac{4}{9} )) \times 4 \\ \\ = \frac{61}{3} - ( \frac{1}{4} + \frac{4}{1} ) \times 4 \\ \\ = \frac{61}{3} - ( \frac{16 + 1}{4} ) \times 4 \\ \\ = \frac{61}{3} - \frac{17}{4} \times \frac{4}{1} \\ \\ = \frac{61}{3} - \frac{17}{1} \\ \\ = \frac{61 - 51}{3} \\ \\ = \frac{10}{3} \\ \\ = 3 \frac{1}{3}$