5. In $\triangle A B C$, $A B=1, A C=2, B-C=\frac{2 \pi}{3}$, then the area of $\triangle A B C$ is $\qquad$
Given, construct $B D=C D=x \Rightarrow A D=2-x$, then $\angle A B D=\frac{2 \pi}{3} \Rightarrow(2-x)^{2}=x^{2}+1+x$ $\Rightarrow x=\frac{3}{5}, 2-x=\frac{7}{5}$. In $\triangle A B D$, $\cos A=\frac{25+49-9}{2 \cdot 5 \cdot 7}=\frac{13}{14} \Rightarrow \sin A=\frac{3 \sqrt{3}}{14} \Rightarrow S_{\triangle A B C}=\frac{3 \sqrt{3}}{14}$.