Question

(1/6а+6,5)-(2/7/9а+3/1/3)

Answer 1

Пошаговое объяснение:

Решение на фото ____---_____

Answer 2

To simplify the expression (1/6a + 6.5) - (2/(7/9a + 3/(1/3)), we'll first deal with the fractions and then combine like terms:

1. For the first fraction, 1/6a, you can leave it as is because it doesn't have a common denominator with the other fractions.

2. For the second fraction, 2/(7/9a + 3/(1/3)), let's find a common denominator. The denominator of the first term is 9a, and the denominator of the second term is 3. The common denominator will be 9a.

Now, we need to adjust the second fraction to have the common denominator:

2/(7/9a + 3/(1/3)) = 2/(7/9a + 9/3)

3. Now, let's combine the fractions:

(1/6a + 6.5) - (2/(7/9a + 9/3))

4. Combine the fractions within the second set of parentheses by finding a common denominator:

(1/6a + 6.5) - (2/((7/9a) + 3))

5. Now, we can add the fractions in the second set of parentheses:

(1/6a + 6.5) - (2/((7/9a) + 3)) = (1/6a + 6.5) - (2/((7/9a) + 3))

6. To simplify further, you might want to get a common denominator for the terms in the first set of parentheses:

(1/6a + 6.5) = (1/6a + 39/6)

7. Now that both fractions have a common denominator, you can combine them:

(1/6a + 39/6) - (2/((7/9a) + 3))

8. To subtract the second fraction, you'll need to find a common denominator again. The common denominator will be 6.

(1/6a + 39/6) - (2/((7/9a) + 18/6))

9. Now, combine the fractions:

(1/6a + 39/6) - (2/((7/9a) + 18/6)) = (1/6a + 39/6) - (2/((7/9a + 18)/6))

10. Invert and multiply to divide by a fraction:

(1/6a + 39/6) - (2 * 6 / (7/9a + 18))

11. Simplify:

(1/6a + 39/6) - (12 / (7/9a + 18))

12. Now, we can combine like terms:

(1/6a + 39/6) - (12 / (7/9a + 18)) = (1/6a + 39/6) - (12 / (7/9a + 18))

13. Finally, you can find a common denominator for the fractions in the first set of parentheses:

(1/6a + 39/6) = (1/6a + 13/2)

14. Now, you have:

(1/6a + 13/2) - (12 / (7/9a + 18))

This is the simplified expression, and further simplification isn't possible without knowing the specific value of 'a'.