Question

(1/2-6а+1/27а³-1:1+3а /1+3а+9а²) •2+6а/а​

Answer 1

Ответ:

1/2-3а+1/27а³

Answer 2

Let's simplify the given expression step by step:

$\left( \frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{1 + 3a + 9a^2} \right) \cdot \left( 2 + \frac{6a}{a} \right)$

First, let's deal with the expression inside the parentheses on the left side:

$\frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{1 + 3a + 9a^2}$

To simplify the last term, we can factor the denominator:

$1 + 3a + 9a^2 = (1 + 3a)(1 + 3a)$

Now, rewrite the expression with a common denominator:

$\frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{(1 + 3a)(1 + 3a)}$

Combine the fractions:

$\frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{(1 + 3a)^2}$

Now, let's multiply this expression by the second part of the given expression:

$\left( \frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{(1 + 3a)^2} \right) \cdot \left( 2 + \frac{6a}{a} \right)$

Combine like terms and simplify:

$\left( \frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{(1 + 3a)^2} \right) \cdot \left( 2 + 6 \right)$

$= \left( \frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{(1 + 3a)^2} \right) \cdot 8$

Now distribute the 8:

$= 8 \left( \frac{1}{2} - 6a + \frac{1}{27a^3} - \frac{1}{(1 + 3a)^2} \right)$

$= 4 - 48a + \frac{8}{27a^3} - \frac{8}{(1 + 3a)^2}$

So, the simplified expression is:

$4 - 48a + \frac{8}{27a^3} - \frac{8}{(1 + 3a)^2}$

Answer 3

Let's simplify the expression step by step:

$\frac{1}{2} - \frac{6a}{1} + \frac{1}{27a^3} - \frac{1}{1} + \frac{3a}{1+3a+9a^2} \times 2 + \frac{6a}{a}$

First, let's simplify the terms individually:

1. $\frac{1}{2} - 6a + \frac{1}{27a^3} - 1$

Combine the numerical terms:

$-\frac{11}{2} - 6a + \frac{1}{27a^3}$

2. $\frac{3a}{1+3a+9a^2} \times 2$

Combine the numerator:

$\frac{6a}{1+3a+9a^2}$

3. $\frac{6a}{a}$

This simplifies to $6$.

Now, let's put it all together:

$-\frac{11}{2} - 6a + \frac{1}{27a^3} + \frac{6a}{1+3a+9a^2} + 6$

This expression is now simplified, but if you have specific values for $a$, you can substitute them in to get a numerical result.