Question 81·Hard·Advanced Math

The expressions form an identity, so they agree at every allowed input. Choose $x=1$: it isn't excluded, and the numerator becomes $1+b+c$. Substitute into both sides: $\frac{1+b+c}{1+3}=1+4+\frac{9}{1+3}$. Choosing $x=1$ turns the unknown coefficients into their sum.

Type the known right-hand side as $h(x)=x+4+\frac{9}{x+3}$, then type $h(1)$. Desmos shows $7.25$, so your equation is $\frac{1+b+c}{4}=7.25$. Keep the $x$ in $x+4$: at $x=1$, that part is $5$, not $4$.

Multiplying the known side by the denominator $4$ recovers $1+b+c$. Then subtract the fixed $1$ to leave the requested sum. Type $4h(1)-1$; Desmos shows $28$. So $b+c=28$. Choice C.