Question

помогите пожалуйста ,вычистите : 5 7/12 + 1 1/3, 6 3/50 - 3 17/75 , 5 6/55 + 3 9/44 - 4 7/22, 2 1/7*2 4/5, 2 1/16:2 5/14, 3 1/6:2 1/9*1 5/9, 5:`5/6+1 2/7*5 5/6- 4 3/4

Answer 1

Ответ:

1) 5 7/12 + 1 1/3 = 5 7/12 + 1 4/12 = 6 11/12;

2) 6 3/50 - 3 17/75 = 6 9/150 - 3 34/150 = 5 159/150 - 3 34/150 = 2 125/150 = 2 5/6;

3) 5 6/55 + 3 9/44 - 4 7/22 = 5 6/55 + 3 9/44 - 4 14/44 = 5 6/55 - 1 5/44 = 5 24/220 - 1 25/220 = 4 244/220 - 1 25/220 = 3 219/220.

Answer 2

Конечно, я помогу вам решить эти математические выражения. Вот решения для каждого из них:

1. $5 \frac{7}{12} + 1 \frac{1}{3}$:

$$5 \frac{7}{12} + 1 \frac{1}{3} = \frac{5 \cdot 12 + 7}{12} + \frac{1 \cdot 3 + 1}{3} = \frac{67}{12} + \frac{4}{3} = \frac{67}{12} + \frac{16}{12} = \frac{83}{12} = 6 \frac{11}{12}.$$

2. $6 \frac{3}{50} - 3 \frac{17}{75}$:

$$6 \frac{3}{50} - 3 \frac{17}{75} = \frac{6 \cdot 50 + 3}{50} - \frac{3 \cdot 75 + 17}{75} = \frac{303}{50} - \frac{236}{75} = \frac{303 \cdot 3}{50 \cdot 3} - \frac{236 \cdot 2}{75 \cdot 2} = \frac{909}{150} - \frac{472}{150} = \frac{437}{150} = 2 \frac{137}{150}.$$

3. $5 \frac{6}{55} + 3 \frac{9}{44} - 4 \frac{7}{22}$:

$$5 \frac{6}{55} + 3 \frac{9}{44} - 4 \frac{7}{22} = \frac{5 \cdot 55 + 6}{55} + \frac{3 \cdot 44 + 9}{44} - \frac{4 \cdot 22 + 7}{22} = \frac{281}{55} + \frac{135}{44} - \frac{99}{22} = \frac{281 \cdot 4}{55 \cdot 4} + \frac{135 \cdot 5}{44 \cdot 5} - \frac{99 \cdot 2}{22 \cdot 2} = \frac{1124}{220} + \frac{675}{220} - \frac{198}{44} = \frac{1124 + 675 - 396}{220} = \frac{1403}{220} = 6 \frac{143}{220}.$$

4. $2 \frac{1}{7} \times 2 \frac{4}{5}$:

$$2 \frac{1}{7} \times 2 \frac{4}{5} = \frac{2 \cdot 7 + 1}{7} \times \frac{2 \cdot 5 + 4}{5} = \frac{15}{7} \times \frac{14}{5} = \frac{15 \cdot 14}{7 \cdot 5} = \frac{210}{35} = 6.$$