Cho số phức z thỏa mãn $\left| z \right|=1$. Gọi M, m lần lượt là...

Ta có: $\left| z \right|=1\Rightarrow {{x}^{2}}=1-{{y}^{2}}$ với $y\in \left[ -1;1 \right]$.
$\begin{aligned}
& P=\left| z+i \right|+2\left| z-2i \right|=\sqrt{{{x}^{2}}+{{\left( y+1 \right)}^{2}}}+2\sqrt{{{x}^{2}}+{{\left( y-2 \right)}^{2}}}=\sqrt{1-{{y}^{2}}+{{\left( y+1 \right)}^{2}}}+2\sqrt{1-{{y}^{2}}+{{\left( y-2 \right)}^{2}}} \\
& \ \ \ =\sqrt{2+2y}+2\sqrt{5-4y}=\sqrt{2\left( 1+y \right)}+2\sqrt{5-4y} \\
\end{aligned}$
Xét hàm số $g\left( x \right)=\sqrt{2\left( 1+y \right)}+2\sqrt{5-4y}$ với $y\in \left[ -1;1 \right]$.
$\begin{aligned}
& g'\left( y \right)=\dfrac{1}{\sqrt{2\left( 1+y \right)}}-\dfrac{4}{\sqrt{5-4y}}. \\
& g'\left( y \right)=0\Leftrightarrow y=-\dfrac{3}{4}. \\
& g\left( -1 \right)=6;g\left( 1 \right)=4;g\left( -\dfrac{3}{4} \right)=\dfrac{9\sqrt{2}}{2}. \\
\end{aligned}$
Do đó $\max P=M=\dfrac{9\sqrt{2}}{2};\min P=m=4$. Suy ra $E=\dfrac{49}{2}$.