[tex]\left(\frac{2}{x^2-4}+\frac{1}{2x-x^2}\right):\frac{1}{x^2+4x+4}[/tex]

Предыдущий
вопрос Следующий
вопрос

The given expression can be simplified by finding the common denominator and adding the terms inside the parentheses.

First, let's find the common denominator for the terms inside the parentheses:

The denominator of the first term is x^2 - 4, which can be factored as (x+2)(x-2).The denominator of the second term is 2x - x^2, which can be rewritten as -x^2 + 2x.

Now, the common denominator for both terms is (x+2)(x-2)(-x^2 + 2x).

Next, rewrite each term with the common denominator and then add them together:

(\frac{2}{x^2-4}) can be rewritten as (\frac{2}{(x+2)(x-2)}).(\frac{1}{2x-x^2}) can be rewritten as (\frac{1}{(-x^2+2x)}).

So, the expression inside the parentheses becomes
[\frac{2}{(x+2)(x-2)} + \frac{1}{(-x^2 + 2x)}]

Now, we multiply the first term by (\frac{(x+2)}{(x+2)}) to get a common denominator
[\frac{2(x+2)}{(x+2)(x-2)} + \frac{1}{(-x^2 + 2x)}]

Now we have
[\frac{2(x+2)}{(x+2)(x-2)} + \frac{1}{(-x^2 + 2x)}:\frac{1}{x^2+4x+4}]

To divide by a fraction, we multiply by its reciprocal
[\frac{2(x+2)}{(x+2)(x-2)} + \frac{1}{(-x^2 + 2x)}\cdot\frac{x^2+4x+4}{1}]

Simplifying further, we get
[\frac{2(x+2)}{(x+2)(x-2)} + \frac{(x^2 + 4x + 4)}{(-x^2 + 2x)}
[\frac{2(x+2)}{(x+2)(x-2)} + \frac{(x+2)(x+2)}{(-x)(x-2)}]

At this step, the numerator and denominator in the second term can cancel out, leading to
[2(x+2) + (x+2)
[2x + 4 + x + 2
[3x + 6]

Therefore, the simplified expression is (3x + 6), assuming x is not equal to 2 or 0.