Cho ${{z}_{1}},{{z}_{2}}$ là hai nghiệm phức của phương trình...

Ta có $\left| z\left[ \left( 2+i \right)\left| z \right|-\left( 1-2i \right) \right] \right|=\left| 1+3i \right|\Rightarrow \left| z \right|.\left| \left( 2+i \right)\left| z \right|-\left( 1-2i \right) \right|=\sqrt{10}$
$\Rightarrow \left| z \right|.\left| \left( 2\left| z \right|-1 \right)+\left( \left| z \right|+2 \right)i \right|=\sqrt{10}\Rightarrow {{\left| z \right|}^{2}}.\left[ {{\left( 2\left| z \right|-1 \right)}^{2}}+{{\left( \left| z \right|+2 \right)}^{2}} \right]=10$
$\Leftrightarrow 5{{\left| z \right|}^{4}}+5{{\left| z \right|}^{2}}-10=0\Leftrightarrow \left| z \right|=1.$
Giả sử ${{z}_{1}}={{x}_{1}}+{{y}_{1}}i$ và ${{z}_{2}}={{x}_{2}}+{{y}_{2}}i\left( {{x}_{1}},{{y}_{1}},{{x}_{2}},{{y}_{2}}\in \mathbb{R} \right)$
Ta có $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1\Rightarrow x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=1$
Bài ra $\left| {{z}_{1}}-{{z}_{2}} \right|=1\Rightarrow \left| {{x}_{1}}-{{x}_{2}}+\left( {{y}_{1}}-{{y}_{2}} \right)i \right|=1\Rightarrow {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}=1$
$\Rightarrow 2-2{{x}_{1}}{{x}_{2}}-2{{y}_{1}}{{y}_{2}}=1\Rightarrow {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}=\dfrac{1}{2}.$
Vậy $P=\left| 2{{z}_{1}}+3{{z}_{2}} \right|=\left| 2\left( {{x}_{1}}+{{y}_{1}}i \right)+3\left( {{x}_{2}}+{{y}_{2}}i \right) \right|=\sqrt{{{\left( 2{{x}_{1}}+3{{x}_{2}} \right)}^{2}}+{{\left( 2{{y}_{1}}+3{{y}_{2}} \right)}^{2}}}$
$=\sqrt{4\left( x_{1}^{2}+y_{1}^{2} \right)+9\left( x_{2}^{2}+y_{2}^{2} \right)+12\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)}=\sqrt{4+9+6}=\sqrt{19}.$