Question

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Answer 1

$\dfrac{q}{q^2+5q+6}+\dfrac{1}{q^2+3q+2}=\dfrac{q}{q^2+3q+2q+6}+\dfrac{1}{q^2+q+2q+2}\\\\=\dfrac{q}{q(q+3)+2(q+3)}+\dfrac{1}{q(q+1)+2(q+1)}\\\\=\dfrac{q}{(q+3)(q+2)}+\dfrac{1}{(q+1)(q+2)}\\\\=\dfrac{q(q+1)}{(q+1)(q+2)(q+3)}+\dfrac{1(q+3)}{(q+1)(q+2)(q+3)}\\\\=\dfrac{q^2+q+q+3}{(q+1)(q+2)(q+3)}=\dfrac{q^2+2q+3}{(q+1)(q+2)(q+3)}\to B.$