Question

(n)/(n+3)+(4n)/(9-n^2)+(12)/(n^2-9)

Answer 1

Вот решение, ечли вам так можно

Answer 2

To simplify the expression:

(n)/(n+3) + (4n)/(9-n^2) + (12)/(n^2-9)

Let's start by factoring the denominators:

n + 3 = (n + 3) 9 - n^2 = -(n^2 - 9) = -(n + 3)(n - 3) n^2 - 9 = (n - 3)(n + 3)

Now, let's rewrite the expression using the factored denominators:

(n)/(n + 3) + (4n)/(-(n + 3)(n - 3)) + (12)/((n - 3)(n + 3))

Next, we can find a common denominator for all the terms, which is (n - 3)(n + 3):

[(n)(n - 3)(n + 3)]/[(n + 3)(n - 3)] + [(4n)(n + 3)]/[(n + 3)(n - 3)] + [12]/[(n - 3)(n + 3)]

Now, we can combine the numerators:

[n^2 - 3n + 4n^2 + 12 + 12(n^2 - 9)]/[(n + 3)(n - 3)]

Simplifying further:

[5n^2 + 12n^2 - 108n + 12]/[(n + 3)(n - 3)]

Combining like terms in the numerator:

[17n^2 - 108n + 12]/[(n + 3)(n - 3)]

Therefore, the simplified form of the expression is:

(17n^2 - 108n + 12)/[(n + 3)(n - 3)]