Cho số phức $z=a+bi\left( a,b\in \mathbb{R} \right)$ thỏa mãn...

Ta có: $\left| z-8 \right|i+\left| z-6i \right|=5\left( 1+i \right)\Leftrightarrow \left[ \begin{aligned}
& \left| z-8 \right|=5 \\
& \left| z-6i \right|=5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( a-8 \right)}^{2}}+{{b}^{2}}=25 \\
& {{a}^{2}}+{{\left( b-6 \right)}^{2}}=25 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& -16a+12b+28=0 \\
& {{a}^{2}}+{{b}^{2}}-12b=-11 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 3b=4a-7 \\
& {{\left( \dfrac{3b+7}{4} \right)}^{2}}+{{b}^{2}}-12b=-11 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 3b=4a-7 \\
& 25{{\left( b-3 \right)}^{2}}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=4 \\
& b=3 \\
\end{aligned} \right.$
Do đó $P=a+b=7$.