Answer:
D
Step-by-step explanation:
[tex]\frac{\frac{b^{2}+5b+6}{3bc}}{\frac{b^{2}-9}{6bc}}\\\\=\frac{b^{2}+5b+6}{3bc}*\frac{6bc}{b^{2}-9}\\\\=\frac{(b+3)*(b+2)}{3bc}*\frac{6bc}{b^{2}-3^{2}}\\\\=\frac{(b+3)*(b+2)}{3bc}*\frac{6bc}{(b+3)(b-3)}\\\\=\frac{(b+2)*2}{(b-3)}\\\\=\frac{2(b+2)}{(b-3)}\\[/tex]
Hint: b
²+ 5b + 6 = b² + 3b + 2b + 2*3
= b*(b + 3) + 2*(b + 3)
= (b + 3) (b + 2)
x² - y² = (x + y)(x - y)
b² - 9 = b² - 3² = (b + 3) (b -3)
