Xét hai số phức ${{z}_{1}}, {{z}_{2}}$ thỏa mãn các điêò kiện...

Gọi ${{z}_{1}}=a+bi, {{z}_{2}}=c+di \left( a,b,c,d\in \mathbb{R} \right)$
Ta có:
$\left| {{z}_{1}} \right|=2\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}=2\Leftrightarrow {{a}^{2}}+{{b}^{2}}=4$
$\left| {{z}_{2}} \right|=\sqrt{3}\Leftrightarrow \sqrt{{{c}^{2}}+{{d}^{2}}}=\sqrt{3}\Leftrightarrow {{c}^{2}}+{{d}^{2}}=3$
$\begin{aligned}
& \left| {{z}_{1}}+{{z}_{2}} \right|=\sqrt{5}\Leftrightarrow \sqrt{{{\left( a+c \right)}^{2}}+{{\left( b+d \right)}^{2}}}=\sqrt{5}\Leftrightarrow \left( {{a}^{2}}+{{b}^{2}} \right)+\left( {{c}^{2}}+{{d}^{2}} \right)+2\left( ac+bd \right)=5 \\
& \Rightarrow ac+bd=-1 \\
\end{aligned}$
Suy ra:
$\left| 3{{z}_{1}}-{{z}_{2}} \right|=\sqrt{{{\left( 3a-c \right)}^{2}}+{{\left( 3b-d \right)}^{2}}}=\sqrt{9\left( {{a}^{2}}+{{b}^{2}} \right)+\left( {{c}^{2}}+{{d}^{2}} \right)-6\left( ac+bd \right)}=3\sqrt{5}$
Khi đó:
$\begin{aligned}
& P=\left| 3{{z}_{1}}-{{z}_{2}}-10+5i \right|+2 \\
& =\left| \left( 3{{z}_{1}}-{{z}_{2}} \right)+\left( -10+5i \right) \right|+2\ge \left| -10+5i \right|-\left| 3{{z}_{1}}-{{z}_{2}} \right|+2=5\sqrt{5}-3\sqrt{5}+2=2+2\sqrt{5} \\
\end{aligned}$