Question

2:1/4+((1 1/2+2 2/3)÷3 3/4-2/3)÷8/89=​

Answer 1

$2:\frac{1}{4}+((1 \frac{1}{2}+2\frac{2}{3}) \div3 \frac{3}{4}-\frac{2}{3})\div\frac{8}{89}=2\cdot 4+((\frac{3}{2}+\frac{8}{3})\div \frac{15}{4}-\frac{2}{3})\cdot \frac{89}{8}= \\ \\ =8+ ((\frac{9+16}{6})\cdot \frac{4}{15}-\frac{2}{3})\cdot \frac{89}{8} = 8+(\frac{25}{6}\cdot \frac{4}{15}-\frac{2}{3})\cdot \frac{89}{8} = 8+(\frac{5\cdot 2}{3\cdot 3}-\frac{2}{3})\cdot \frac{89}{8}=8+(\frac{10}{9}-\frac{2}{3})\cdot \frac{89}{8}=$

$=8+\frac{4}{9}\cdot \frac{89}{8}=8+\frac{89}{9\cdot 2}=8+\frac{89}{18}=\frac{144+89}{18}=\frac{233}{18}$

Answer 2

Ответ:

Вот

Пошаговое объяснение:

Решила так, как смогла

Answer 3

Let's break down the expression step by step:

$\frac{2}{1/4} + \left( \frac{1}{1/2} + \frac{2}{2/3} \right) \div \left( \frac{3}{3/4} - \frac{2}{3} \right) \div \frac{8}{89}$

1. Start by simplifying the fractions in the expression:

   $\frac{2}{1/4} = 2 \times \frac{4}{1} = 8$

   $\frac{1}{1/2} = 1 \times \frac{2}{1} = 2$

   $\frac{2}{2/3} = 2 \times \frac{3}{2} = 3$

   $\frac{3}{3/4} = 3 \times \frac{4}{3} = 4$

2. Substitute these values back into the expression:

   $8 + \left( 2 + 3 \right) \div \left( 4 - \frac{2}{3} \right) \div \frac{8}{89}$

3. Continue simplifying:

   $8 + 5 \div \left( 4 - \frac{2}{3} \right) \div \frac{8}{89}$

   Now, find a common denominator for the subtraction:

   $\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}$

   $8 + 5 \div \left( 4 - \frac{6}{9} \right) \div \frac{8}{89}$

   $8 + 5 \div \left( \frac{36}{9} - \frac{6}{9} \right) \div \frac{8}{89}$

   $8 + 5 \div \left( \frac{30}{9} \right) \div \frac{8}{89}$

   $8 + 5 \div \frac{30}{9} \div \frac{8}{89}$

4. Simplify further:

   $8 + 5 \div \frac{30}{9} \div \frac{8}{89}$

   $8 + 5 \div \frac{30}{9} \times \frac{89}{8}$

   $8 + 5 \times \frac{89}{8} \div \frac{30}{9}$

   $8 + \frac{445}{8} \div \frac{30}{9}$

   Now, convert the division to multiplication by taking the reciprocal:

   $8 + \frac{445}{8} \times \frac{9}{30}$

   $8 + \frac{1335}{40}$