Answer:
[tex]\frac{1}{(x+1)(x-2)}[/tex]
Step-by-step explanation:
[tex]\frac{x+2}{4x^2 + 5x + 1} \ \times \ \frac{4x+1}{x^2-4}\\\\=\frac{x+2}{4x^2 + 4x + x + 1} \ \times \ \frac{4x+1}{x^2-2^2}\\\\=\frac{x+2}{4x(x + 1) + 1( x + 1)} \ \times \ \frac{4x+1}{(x - 2)(x + 2)} \ \ \ \ \ \ \ \ [ \ (a^2 - b^2 = (a-b)(a+b) \ ]\\\\\\=\frac{x+2}{(4x + 1)(x+1)} \ \times \ \frac{4x+1}{(x-2)(x+2)}\\\\=\frac{1}{(x+1)} \ \times \ \frac{1}{(x-2)}\\\\= \frac{1}{(x+1)(x-2)}[/tex]
Answer:
D. 1/(x+1)(x-2)
Step-by-step explanation:
i looked it up on a simplifier :)
