9. Given $\triangle A B C$ satisfies $A B=1, A C=2, \cos B+\sin C=1$, find the length of side $B C$.

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From the Law of Sines, we get $\frac{b}{\sin B}=\frac{c}{\sin C} \Rightarrow \sin B=2 \sin C$, thus $\cos B+\frac{1}{2} \sin B=1$. Substituting $\sin ^{2} B+\cos ^{2} B=1$ yields $\sin B=\frac{4}{5}, \cos B=\frac{3}{5}, \sin C=\frac{2}{5}, \cos C=\frac{\sqrt{21}}{5}$.
Therefore, $\cos A=-\cos (B+C)=\frac{4}{5} \cdot \frac{2}{5}-\frac{3}{5} \cdot \frac{\sqrt{21}}{5}=\frac{8-3 \sqrt{21}}{25}$.
Hence, $a^{2}=1+4+2 \times 1 \times 2 \times \frac{3 \sqrt{21}-8}{25}=\frac{12 \sqrt{21}+93}{25} \Rightarrow B C=\frac{2 \sqrt{21}+3}{5}$.