Answer :

Sure, let's solve this step by step. We are given the following operation to perform:

[tex]\[ \left(2 m^2+2\right) \div \frac{m^4-1}{m^2+8m-9} \][/tex]

First, let's rewrite the division as multiplication by the reciprocal:

[tex]\[ \left(2 m^2+2\right) \div \frac{m^4-1}{m^2+8m-9} = \left(2 m^2+2\right) \times \frac{m^2 + 8m - 9}{m^4 - 1} \][/tex]

Next, let's simplify the given expressions. Consider the numerator \(2m^2 + 2\):

[tex]\[ 2m^2 + 2 = 2(m^2 + 1) \][/tex]

And for the denominator \(\frac{m^4-1}{m^2+8m-9}\):

First, factorize \(m^4 - 1\):
[tex]\[ m^4 - 1 = (m^2 - 1)(m^2 + 1) = (m - 1)(m + 1)(m^2 + 1) \][/tex]

Next, factorize \(m^2 + 8m - 9\):
[tex]\[ m^2 + 8m - 9 = (m + 9)(m - 1) \][/tex]

Now the expression becomes:

[tex]\[ \left(2(m^2 + 1)\right) \times \frac{(m + 9)(m - 1)}{(m - 1)(m + 1)(m^2 + 1)} \][/tex]

We can then cancel out the common factors \((m^2 + 1)\) and \((m - 1)\):

[tex]\[ 2 \times \frac{m + 9}{m + 1} \][/tex]

Therefore, the simplified result is:

[tex]\[ \frac{2(m + 9)}{m + 1} \][/tex]

Thus, the final answer is:

[tex]\[ 2\left(\frac{m + 9}{m + 1}\right) = \boxed{2\frac{m + 9}{m + 1}} \][/tex]