Cho z1;z2 là các số phức thỏa mãn điều kiện

* Hướng dẫn giải

Đặt $z_1=a+bi$; $z_2=c+di$ với a,b,c,d$\in \mathrm{ℝ}$

Ta có 

$$\begin{gathered} P=\left|z_1+z_2\right| \\ =\sqrt{{\left(a+c\right)}^2+{\left(b+d\right)}^2} \\ =\sqrt{a^2+b^2+c^2+d^2+2ac+2bd} \end{gathered}$$

Theo đề ta có 

$$\begin{gathered} \left\{\begin{array}{l}\left|z_1-2i\right|=\sqrt{2}\left|i{z}_1+1\right|\\ \left|z_2-2i\right|=\sqrt{2}\left|i{z}_2+1\right|\\ \left|z_1-z_2\right|=1\end{array}\right. \\ \iff \left\{\begin{array}{l}\sqrt{a^2+{\left(b-2\right)}^2}\\ \sqrt{c^2+{\left(d-2\right)}^2}\\ \sqrt{{\left(a-c\right)}^2+{\left(b-d\right)}^2}=1\end{array}\right. \\ \iff \left\{\begin{array}{l}=\sqrt{2}\sqrt{{\left(1-b\right)}^2+a^2}\\ =\sqrt{2}\sqrt{{\left(1-d\right)}^2+c^2}\\ =1\end{array}\right. \\ \iff \left\{\begin{array}{l}{a}^2+b^2=2\\ c^2+d^2=2\\ a^2+b^2+c^2+d^2-2ac-2bd=1\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}{a}^2+b^2+c^2+d^2=4\\ 2ac+2bd=3\end{array}\right. \end{gathered}$$

Suy ra

$$\begin{gathered} P=\sqrt{a^2+b^2+c^2+d^2} \\ +\sqrt{2ac+2bd}=\sqrt{7} \end{gathered}$$

Đáp án B