Ta chọn câu D
Ta có: \(z^{2} -6z+13=0\Leftrightarrow \left[\begin{array}{l} {z=3+2i} \\ {z=3-2i} \end{array}\right. \)
TH 1: \(z=3+2i, z+\frac{6}{z+i} =3+2i+\frac{6}{3+2i+i} =3+2i+1-i=4+i\)
Suy ra \(\left|z+\frac{6}{z+i} \right|=\left|4+i\right|=\sqrt{4^{2} +1^{2} } =\sqrt{17} \)
TH 2: \(z=3-2i, z+\frac{6}{z+i} =3-2i+\frac{6}{3-2i+i} =3-2i+\frac{9}{5} +\frac{3}{5} i=\frac{24}{5} -\frac{7}{5} i\)
Suy ra \(\left|z+\frac{6}{z+i} \right|=\left|\frac{24}{5} -\frac{7}{5} i\right| =\sqrt{\left(\frac{24}{5} \right)^{2} +\left(-\frac{7}{5} \right)^{2} } =5.\)