# Simplify $(\frac{x+5}{x^2-81} + \frac{x+7}{x^2-18x+81}):(\frac{x+3}{x-9})^2 + \frac{7+x}{9+x}$ [closed]

Simplify $$\bigg(\frac{x+5}{x^2-81} + \frac{x+7}{x^2-18x+81}\bigg):\bigg(\frac{x+3}{x-9}\bigg)^2 + \frac{7+x}{9+x}.$$

I was told that the right answer is $\displaystyle\frac{3x+25}{x+9}$, but the best what I can get is $\displaystyle\frac{2x^2+12x-18}{(x+9)(x+3)^2}$.

I would be very thankful if someone could help me:)

$$(\frac{x+5}{x^2-81} + \frac{x+7}{x^2-18x+81}):(\frac{x+3}{x-9})^2 + \frac{7+x}{9+x}=$$ $$=\left(\frac{x+5}{(x-9)(x+9)}+\frac{x+7}{(x-9)^2}\right)\cdot\frac{(x-9)^2}{(x+3)^2}+\frac{x+7}{x+9}=$$ $$=\frac{x^2-4x-45+x^2+16x+63}{(x-9)^2(x+9)}\cdot\frac{(x-9)^2}{(x+3)^2}+\frac{x+7}{x+9}=$$ $$=\frac{2(x+3)^2}{(x-9)^2(x+9)}\cdot\frac{(x-9)^2}{(x+3)^2}+\frac{x+7}{x+9}=1$$

$$\frac{x+5}{x^2-81} + \frac{x+7}{x^2-18x+81} \equiv \frac{x+5}{(x-9)(x+9)} + \frac{x+7}{(x-9)^2}$$

$$\equiv \frac{(x+5)(x-9)+(x+7)(x+9)}{(x-9)^2(x+9)}$$ $$\equiv \frac{x^2-4x-45+x^2+16x+63}{(x-9)^2(x+9)}$$ $$\equiv \frac{2x^2+12x+18}{(x-9)^2(x+9)}$$ $$\equiv \frac{2(x^2+6x+9)}{(x-9)^2(x+9)} \equiv \frac{2(x+3)^2}{(x-9)^2(x+9)}$$

Now, we divide by $(\frac{x+3}{x-9})^2$:

$$\require{cancel} \frac{2\cancel{(x+3)^2}}{\cancel{(x-9)^2}(x+9)} \times \frac{\cancel{(x-9)^2}}{\cancel{(x+3)^2}} \equiv \frac{2}{x+9}$$

And adding $\frac{7+x}{9+x}$:

$$\frac{2}{x+9}+\frac{x+7}{x+9} \equiv \frac{x+9}{x+9} \equiv 1$$

Note that as we've used fractions with denominators of $x+9$ and $x-9$, the result is undefined (rather than $1$) for $x = \pm 9$. Wolfram Alpha confirms that this is the correct answer, rather than $\frac{3x+25}{x+9}$.