Cho số phức $z=a+bi(a,b\in \mathbb{R})$ thỏa mãn $\left| z-8...

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