\(NaX+AgNO_3\rightarrow AgX\downarrow+NaNO_3\\ n_X=n_{NaX}=n_{AgX}=\dfrac{18,797-10,291}{108-23}=0,1\left(mol\right)\\ M_{NaX}=23+M_X=\dfrac{10,291}{0,1}=102,91\\ \Leftrightarrow M_X=79,91\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Brom\left(Br\right).Có.2.đồng.vị:^{79}Br,^{81}Br\\ Ta.có:\overline{M}_{Br}=79,91\\ \Leftrightarrow79.x_1+81x_2=79,91\left(1\right)\\ Và:x_1+x_2=100\%\left(2\right)\\ \Rightarrow x_1=54,5\%;x_2=45,5\%\\ \Rightarrow Chọn.A\)