Answer:
[tex] \frac{1}{3x+1} [/tex]
Explanation:
1) Expression given:
[tex] \frac{ \frac{x+2}{x^3+2x^2-9x-18} }{ \frac{3x+1}{x^2-9} } [/tex]
2) Factor the denominator of the first fraction:
[tex]x^3+2x^2-9x-18=(x^3+2x^2)-(9x+18)=x^2(x+2)-9(x+2)=[/tex]
[tex](x+2)(x^2-9)[/tex]
3) Rewrite the expression
[tex] \frac{ \frac{x+2}{(x+2)(x^2-9)} }{ \frac{3x+1}{x^2-9} } [/tex]
4) In the upper fraction x+2 in the numerator cancels out with x + 2 in the denominator.
[tex] \frac{ \frac{1}{x^2-9} }{ \frac{3x+1}{x^2-9} } [/tex]
5) x² - 9 in both denominators cancel out each other, and the final result is:
[tex] \frac{1}{3x+1} [/tex]
